This document provides a rigorous treatment of modulus-argument form, De Moivre’s theorem, roots of Complex numbers, loci in the Argand diagram, and Euler’s formula.
Difficult to prove in real analysis become straightforward when extended to the complex plane. Any non-zero complex number z = x + i y z = x + iy z = x + i y modulus-argument form (polar form):
z = r ( cos θ + i sin θ ) = r c i s θ z = r(\cos\theta + i\sin\theta) = r\,\mathrm{cis}\,\theta z = r ( cos θ + i sin θ ) = r cis θ
Where r = ∣ z ∣ = x 2 + y 2 r = |z| = \sqrt{x^2 + y^2} r = ∣ z ∣ = x 2 + y 2 θ = arg ( z ) \theta = \arg(z) θ = arg ( z )
The argument is multi-valued: arg ( z ) = θ + 2 k π \arg(z) = \theta + 2k\pi arg ( z ) = θ + 2 k π k ∈ Z k \in \mathbb{Z} k ∈ Z principal Argument A r g ( z ) \mathrm{Arg}(z) Arg ( z ) − π < A r g ( z ) ≤ π -\pi \lt \mathrm{Arg}(z) \leq \pi − π < Arg ( z ) ≤ π
If z 1 = r 1 ( cos θ 1 + i sin θ 1 ) z_1 = r_1(\cos\theta_1 + i\sin\theta_1) z 1 = r 1 ( cos θ 1 + i sin θ 1 ) z 2 = r 2 ( cos θ 2 + i sin θ 2 ) z_2 = r_2(\cos\theta_2 + i\sin\theta_2) z 2 = r 2 ( cos θ 2 + i sin θ 2 )
z 1 z 2 = r 1 r 2 ( cos ( θ 1 + θ 2 ) + i sin ( θ 1 + θ 2 ) ) z_1 z_2 = r_1 r_2\bigl(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\bigr) z 1 z 2 = r 1 r 2 ( cos ( θ 1 + θ 2 ) + i sin ( θ 1 + θ 2 ) )
z 1 z 2 = r 1 r 2 ( cos ( θ 1 − θ 2 ) + i sin ( θ 1 − θ 2 ) ) \frac{z_1}{z_2} = \frac{r_1}{r_2}\bigl(\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)\bigr) z 2 z 1 = r 2 r 1 ( cos ( θ 1 − θ 2 ) + i sin ( θ 1 − θ 2 ) )
Proof. Using the compound angle formulas:
z 1 z 2 = r 1 r 2 ( cos θ 1 cos θ 2 − sin θ 1 sin θ 2 + i ( sin θ 1 cos θ 2 + cos θ 1 sin θ 2 ) ) z_1 z_2 = r_1 r_2\bigl(\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2 + i(\sin\theta_1\cos\theta_2 + \cos\theta_1\sin\theta_2)\bigr) z 1 z 2 = r 1 r 2 ( cos θ 1 cos θ 2 − sin θ 1 sin θ 2 + i ( sin θ 1 cos θ 2 + cos θ 1 sin θ 2 ) )
= r 1 r 2 ( cos ( θ 1 + θ 2 ) + i sin ( θ 1 + θ 2 ) ) ■ = r_1 r_2\bigl(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)\bigr) \quad \blacksquare = r 1 r 2 ( cos ( θ 1 + θ 2 ) + i sin ( θ 1 + θ 2 ) ) ■
This confirms: ∣ z 1 z 2 ∣ = ∣ z 1 ∣ ∣ z 2 ∣ |z_1 z_2| = |z_1||z_2| ∣ z 1 z 2 ∣ = ∣ z 1 ∣∣ z 2 ∣ arg ( z 1 z 2 ) = arg ( z 1 ) + arg ( z 2 ) \arg(z_1 z_2) = \arg(z_1) + \arg(z_2) arg ( z 1 z 2 ) = arg ( z 1 ) + arg ( z 2 )
Problem. Express 1 + i 3 1 − i \dfrac{1 + i\sqrt{3}}{1 - i} 1 − i 1 + i 3
Numerator: 1 + i 3 1 + i\sqrt{3} 1 + i 3 r 1 = 1 + 3 = 2 r_1 = \sqrt{1 + 3} = 2 r 1 = 1 + 3 = 2 θ 1 = arctan ( 3 1 ) = π 3 \theta_1 = \arctan\!\left(\dfrac{\sqrt{3}}{1}\right) = \dfrac{\pi}{3} θ 1 = arctan ( 1 3 ) = 3 π
Denominator: 1 − i 1 - i 1 − i r 2 = 1 + 1 = 2 r_2 = \sqrt{1 + 1} = \sqrt{2} r 2 = 1 + 1 = 2 θ 2 = arctan ( − 1 1 ) = − π 4 \theta_2 = \arctan\!\left(\dfrac{-1}{1}\right) = -\dfrac{\pi}{4} θ 2 = arctan ( 1 − 1 ) = − 4 π
z 1 z 2 = 2 2 ( cos ( π 3 − ( − π 4 ) ) + i sin ( 7 π 12 ) ) \frac{z_1}{z_2} = \frac{2}{\sqrt{2}}\left(\cos\!\left(\frac{\pi}{3} - \left(-\frac{\pi}{4}\right)\right) + i\sin\!\left(\frac{7\pi}{12}\right)\right) z 2 z 1 = 2 2 ( cos ( 3 π − ( − 4 π ) ) + i sin ( 12 7 π ) )
= 2 ( cos 7 π 12 + i sin 7 π 12 ) = \sqrt{2}\left(\cos\frac{7\pi}{12} + i\sin\frac{7\pi}{12}\right) = 2 ( cos 12 7 π + i sin 12 7 π )
Theorem (De Moivre). For any integer n n n
( r ( cos θ + i sin θ ) ) n = r n ( cos n θ + i sin n θ ) \bigl(r(\cos\theta + i\sin\theta)\bigr)^n = r^n\bigl(\cos n\theta + i\sin n\theta\bigr) ( r ( cos θ + i sin θ ) ) n = r n ( cos n θ + i sin n θ )
Proof by induction for n ≥ 0 n \geq 0 n ≥ 0
Base case n = 0 n = 0 n = 0 ( cos θ + i sin θ ) 0 = 1 = cos 0 + i sin 0 (\cos\theta + i\sin\theta)^0 = 1 = \cos 0 + i\sin 0 ( cos θ + i sin θ ) 0 = 1 = cos 0 + i sin 0
Inductive step: Assume true for n = k n = k n = k
( cos θ + i sin θ ) k + 1 = ( cos θ + i sin θ ) k ( cos θ + i sin θ ) (\cos\theta + i\sin\theta)^{k+1} = (\cos\theta + i\sin\theta)^k(\cos\theta + i\sin\theta) ( cos θ + i sin θ ) k + 1 = ( cos θ + i sin θ ) k ( cos θ + i sin θ )
= ( cos k θ + i sin k θ ) ( cos θ + i sin θ ) = (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta) = ( cos k θ + i sin k θ ) ( cos θ + i sin θ )
= cos ( k + 1 ) θ + i sin ( k + 1 ) θ = \cos(k+1)\theta + i\sin(k+1)\theta = cos ( k + 1 ) θ + i sin ( k + 1 ) θ
By the compound angle formulas. True for n = k + 1 n = k + 1 n = k + 1 ■ \blacksquare ■
For negative integers, note that:
( cos θ + i sin θ ) − 1 = cos ( − θ ) + i sin ( − θ ) = cos θ − i sin θ (\cos\theta + i\sin\theta)^{-1} = \cos(-\theta) + i\sin(-\theta) = \cos\theta - i\sin\theta ( cos θ + i sin θ ) − 1 = cos ( − θ ) + i sin ( − θ ) = cos θ − i sin θ
And the result follows by applying the positive case to the reciprocal.
Problem. Find ( 1 + i ) 10 (1 + i)^{10} ( 1 + i ) 10
1 + i = 2 ( cos π 4 + i sin π 4 ) 1 + i = \sqrt{2}\!\left(\cos\dfrac{\pi}{4} + i\sin\dfrac{\pi}{4}\right) 1 + i = 2 ( cos 4 π + i sin 4 π )
( 1 + i ) 10 = ( 2 ) 10 ( cos 5 π 2 + i sin 5 π 2 ) = 32 ( 0 + i ) = 32 i (1 + i)^{10} = (\sqrt{2})^{10}\!\left(\cos\frac{5\pi}{2} + i\sin\frac{5\pi}{2}\right) = 32(0 + i) = 32i ( 1 + i ) 10 = ( 2 ) 10 ( cos 2 5 π + i sin 2 5 π ) = 32 ( 0 + i ) = 32 i
De Moivre’s theorem provides a systematic way to derive multiple-angle formulas.
Example: Expanding ( cos θ + i sin θ ) 3 (\cos\theta + i\sin\theta)^3 ( cos θ + i sin θ ) 3
cos 3 θ + i sin 3 θ = cos 3 θ + 3 i cos 2 θ sin θ − 3 cos θ sin 2 θ − i sin 3 θ \cos 3\theta + i\sin 3\theta = \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta cos 3 θ + i sin 3 θ = cos 3 θ + 3 i cos 2 θ sin θ − 3 cos θ sin 2 θ − i sin 3 θ
Equating real parts:
cos 3 θ = cos 3 θ − 3 cos θ sin 2 θ = 4 cos 3 θ − 3 cos θ \cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta = 4\cos^3\theta - 3\cos\theta cos 3 θ = cos 3 θ − 3 cos θ sin 2 θ = 4 cos 3 θ − 3 cos θ
Equating imaginary parts:
sin 3 θ = 3 cos 2 θ sin θ − sin 3 θ = 3 sin θ − 4 sin 3 θ \sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta = 3\sin\theta - 4\sin^3\theta sin 3 θ = 3 cos 2 θ sin θ − sin 3 θ = 3 sin θ − 4 sin 3 θ
To solve z n = w z^n = w z n = w w = R ( cos α + i sin α ) w = R(\cos\alpha + i\sin\alpha) w = R ( cos α + i sin α )
z = R 1 / n ( cos α + 2 k π n + i sin α + 2 k π n ) , k = 0 , 1 , 2 , … , n − 1 z = R^{1/n}\!\left(\cos\frac{\alpha + 2k\pi}{n} + i\sin\frac{\alpha + 2k\pi}{n}\right), \quad k = 0, 1, 2, \ldots, n-1 z = R 1/ n ( cos n α + 2 k π + i sin n α + 2 k π ) , k = 0 , 1 , 2 , … , n − 1
The n n n w w w R 1 / n R^{1/n} R 1/ n 2 π n \dfrac{2\pi}{n} n 2 π
The sum of all n n n z n = w z^n = w z n = w
Proof. The roots are R 1 / n ω k R^{1/n}\,\omega^k R 1/ n ω k ω = c i s ( 2 π / n ) \omega = \mathrm{cis}(2\pi/n) ω = cis ( 2 π / n ) k = 0 , 1 , … , n − 1 k = 0, 1, \ldots, n-1 k = 0 , 1 , … , n − 1
∑ k = 0 n − 1 ω k = 1 − ω n 1 − ω = 1 − 1 1 − ω = 0 ■ \sum_{k=0}^{n-1}\omega^k = \frac{1 - \omega^n}{1 - \omega} = \frac{1 - 1}{1 - \omega} = 0 \quad \blacksquare ∑ k = 0 n − 1 ω k = 1 − ω 1 − ω n = 1 − ω 1 − 1 = 0 ■
Problem. Find all cube roots of − 8 -8 − 8
− 8 = 8 ( cos π + i sin π ) -8 = 8(\cos\pi + i\sin\pi) − 8 = 8 ( cos π + i sin π )
z k = 8 1 / 3 ( cos π + 2 k π 3 + i sin π + 2 k π 3 ) , k = 0 , 1 , 2 z_k = 8^{1/3}\!\left(\cos\frac{\pi + 2k\pi}{3} + i\sin\frac{\pi + 2k\pi}{3}\right), \quad k = 0, 1, 2 z k = 8 1/3 ( cos 3 π + 2 k π + i sin 3 π + 2 k π ) , k = 0 , 1 , 2
k = 0 k = 0 k = 0 z 0 = 2 ( cos π 3 + i sin π 3 ) = 1 + i 3 z_0 = 2\!\left(\cos\dfrac{\pi}{3} + i\sin\dfrac{\pi}{3}\right) = 1 + i\sqrt{3} z 0 = 2 ( cos 3 π + i sin 3 π ) = 1 + i 3
k = 1 k = 1 k = 1 z 1 = 2 ( cos π + i sin π ) = − 2 z_1 = 2\!\left(\cos\pi + i\sin\pi\right) = -2 z 1 = 2 ( cos π + i sin π ) = − 2
k = 2 k = 2 k = 2 z 2 = 2 ( cos 5 π 3 + i sin 5 π 3 ) = 1 − i 3 z_2 = 2\!\left(\cos\dfrac{5\pi}{3} + i\sin\dfrac{5\pi}{3}\right) = 1 - i\sqrt{3} z 2 = 2 ( cos 3 5 π + i sin 3 5 π ) = 1 − i 3
Check: ( 1 + i 3 ) + ( − 2 ) + ( 1 − i 3 ) = 0 (1 + i\sqrt{3}) + (-2) + (1 - i\sqrt{3}) = 0 ( 1 + i 3 ) + ( − 2 ) + ( 1 − i 3 ) = 0
The n n n z n = 1 z^n = 1 z n = 1
z k = cos 2 k π n + i sin 2 k π n , k = 0 , 1 , … , n − 1 z_k = \cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n}, \quad k = 0, 1, \ldots, n-1 z k = cos n 2 k π + i sin n 2 k π , k = 0 , 1 , … , n − 1
These form a regular n n n
Key property: ∑ k = 0 n − 1 z k = 0 \displaystyle\sum_{k=0}^{n-1} z_k = 0 k = 0 ∑ n − 1 z k = 0 ∏ k = 0 n − 1 z k = ( − 1 ) n − 1 \displaystyle\prod_{k=0}^{n-1} z_k = (-1)^{n-1} k = 0 ∏ n − 1 z k = ( − 1 ) n − 1
∣ z − z 0 ∣ = r |z - z_0| = r ∣ z − z 0 ∣ = r z 0 z_0 z 0 r r r
Proof. If z = x + i y z = x + iy z = x + i y z 0 = a + i b z_0 = a + ib z 0 = a + ib
∣ z − z 0 ∣ = ( x − a ) 2 + ( y − b ) 2 = r ⟹ ( x − a ) 2 + ( y − b ) 2 = r 2 ■ |z - z_0| = \sqrt{(x - a)^2 + (y - b)^2} = r \implies (x - a)^2 + (y - b)^2 = r^2 \quad \blacksquare ∣ z − z 0 ∣ = ( x − a ) 2 + ( y − b ) 2 = r ⟹ ( x − a ) 2 + ( y − b ) 2 = r 2 ■
∣ z − z 1 ∣ = ∣ z − z 2 ∣ |z - z_1| = |z - z_2| ∣ z − z 1 ∣ = ∣ z − z 2 ∣ z 1 z_1 z 1 z 2 z_2 z 2
∣ z − z 0 ∣ < r |z - z_0| \lt r ∣ z − z 0 ∣ < r
∣ z − z 0 ∣ > r |z - z_0| \gt r ∣ z − z 0 ∣ > r
arg ( z − z 0 ) = α \arg(z - z_0) = \alpha arg ( z − z 0 ) = α z 0 z_0 z 0 α \alpha α
α < arg ( z − z 0 ) < β \alpha \lt \arg(z - z_0) \lt \beta α < arg ( z − z 0 ) < β
Problem. Sketch the region defined by ∣ z − 2 ∣ ≤ 3 |z - 2| \leq 3 ∣ z − 2∣ ≤ 3 0 ≤ arg ( z ) ≤ π 4 0 \leq \arg(z) \leq \dfrac{\pi}{4} 0 ≤ arg ( z ) ≤ 4 π
∣ z − 2 ∣ ≤ 3 |z - 2| \leq 3 ∣ z − 2∣ ≤ 3 2 + 0 i 2 + 0i 2 + 0 i arg z = π / 4 \arg z = \pi/4 arg z = π /4
The disc extends from x = − 1 x = -1 x = − 1 x = 5 x = 5 x = 5 arg z = π / 4 \arg z = \pi/4 arg z = π /4 y = x y = x y = x y = x y = x y = x ( x − 2 ) 2 + y 2 = 9 (x-2)^2 + y^2 = 9 ( x − 2 ) 2 + y 2 = 9
( x − 2 ) 2 + x 2 = 9 ⟹ 2 x 2 − 4 x − 5 = 0 ⟹ x = 4 ± 16 + 40 4 = 4 ± 56 4 (x - 2)^2 + x^2 = 9 \implies 2x^2 - 4x - 5 = 0 \implies x = \frac{4 \pm \sqrt{16 + 40}}{4} = \frac{4 \pm \sqrt{56}}{4} ( x − 2 ) 2 + x 2 = 9 ⟹ 2 x 2 − 4 x − 5 = 0 ⟹ x = 4 4 ± 16 + 40 = 4 4 ± 56
The relevant intersection is at x = 1 + 14 2 ≈ 2.87 x = 1 + \dfrac{\sqrt{14}}{2} \approx 2.87 x = 1 + 2 14 ≈ 2.87
Problem. Find the Cartesian equation of the locus ∣ z − 3 + 2 i ∣ = 2 ∣ z + 1 − i ∣ |z - 3 + 2i| = 2|z + 1 - i| ∣ z − 3 + 2 i ∣ = 2∣ z + 1 − i ∣
Let z = x + i y z = x + iy z = x + i y
( x − 3 ) 2 + ( y + 2 ) 2 = 2 ( x + 1 ) 2 + ( y − 1 ) 2 \sqrt{(x - 3)^2 + (y + 2)^2} = 2\sqrt{(x + 1)^2 + (y - 1)^2} ( x − 3 ) 2 + ( y + 2 ) 2 = 2 ( x + 1 ) 2 + ( y − 1 ) 2
( x − 3 ) 2 + ( y + 2 ) 2 = 4 ( x + 1 ) 2 + 4 ( y − 1 ) 2 (x - 3)^2 + (y + 2)^2 = 4(x + 1)^2 + 4(y - 1)^2 ( x − 3 ) 2 + ( y + 2 ) 2 = 4 ( x + 1 ) 2 + 4 ( y − 1 ) 2
x 2 − 6 x + 9 + y 2 + 4 y + 4 = 4 x 2 + 8 x + 4 + 4 y 2 − 8 y + 4 x^2 - 6x + 9 + y^2 + 4y + 4 = 4x^2 + 8x + 4 + 4y^2 - 8y + 4 x 2 − 6 x + 9 + y 2 + 4 y + 4 = 4 x 2 + 8 x + 4 + 4 y 2 − 8 y + 4
3 x 2 + 14 x + 3 y 2 − 12 y − 5 = 0 3x^2 + 14x + 3y^2 - 12y - 5 = 0 3 x 2 + 14 x + 3 y 2 − 12 y − 5 = 0
Completing the square:
3 ( x + 7 3 ) 2 + 3 ( y − 2 ) 2 = 5 + 49 3 + 12 = 100 3 3\!\left(x + \frac{7}{3}\right)^{\!2} + 3\!\left(y - 2\right)^{\!2} = 5 + \frac{49}{3} + 12 = \frac{100}{3} 3 ( x + 3 7 ) 2 + 3 ( y − 2 ) 2 = 5 + 3 49 + 12 = 3 100
( x + 7 3 ) 2 + ( y − 2 ) 2 = 100 9 \left(x + \frac{7}{3}\right)^{\!2} + (y - 2)^2 = \frac{100}{9} ( x + 3 7 ) 2 + ( y − 2 ) 2 = 9 100
This is a circle with centre ( − 7 3 , 2 ) \left(-\dfrac{7}{3}, 2\right) ( − 3 7 , 2 ) 10 3 \dfrac{10}{3} 3 10
Euler’s formula:
e i θ = cos θ + i sin θ \boxed{e^{i\theta} = \cos\theta + i\sin\theta} e i θ = cos θ + i sin θ
This connects the exponential function with trigonometric functions via the imaginary unit.
e i θ = ∑ n = 0 ∞ ( i θ ) n n ! = 1 + i θ + ( i θ ) 2 2 ! + ( i θ ) 3 3 ! + ⋯ e^{i\theta} = \sum_{n=0}^{\infty}\frac{(i\theta)^n}{n!} = 1 + i\theta + \frac{(i\theta)^2}{2!} + \frac{(i\theta)^3}{3!} + \cdots e i θ = ∑ n = 0 ∞ n ! ( i θ ) n = 1 + i θ + 2 ! ( i θ ) 2 + 3 ! ( i θ ) 3 + ⋯
Since i^2 = -1$$i^3 = -i$$i^4 = 1 And this pattern repeats with period 4:
= ( 1 − θ 2 2 ! + θ 4 4 ! − ⋯ ) + i ( θ − θ 3 3 ! + θ 5 5 ! − ⋯ ) = \left(1 - \frac{\theta^2}{2!} + \frac{\theta^4}{4!} - \cdots\right) + i\left(\theta - \frac{\theta^3}{3!} + \frac{\theta^5}{5!} - \cdots\right) = ( 1 − 2 ! θ 2 + 4 ! θ 4 − ⋯ ) + i ( θ − 3 ! θ 3 + 5 ! θ 5 − ⋯ )
= cos θ + i sin θ ■ = \cos\theta + i\sin\theta \quad \blacksquare = cos θ + i sin θ ■
Euler’s identity: Setting θ = π \theta = \pi θ = π
e i π + 1 = 0 e^{i\pi} + 1 = 0 e iπ + 1 = 0
This connects five fundamental constants: e$$i$$\pi$$1 And 0 0 0
Complex exponential form: Any complex number can be written as:
z = r e i θ z = re^{i\theta} z = r e i θ
Where r = ∣ z ∣ r = |z| r = ∣ z ∣ θ = arg ( z ) \theta = \arg(z) θ = arg ( z )
( r e i θ ) n = r n e i n θ \bigl(re^{i\theta}\bigr)^n = r^n e^{in\theta} ( r e i θ ) n = r n e in θ
z 1 z 2 = r 1 r 2 e i ( θ 1 + θ 2 ) z_1 z_2 = r_1 r_2\,e^{i(\theta_1 + \theta_2)} z 1 z 2 = r 1 r 2 e i ( θ 1 + θ 2 )
From Euler’s formula:
cos θ = e i θ + e − i θ 2 \cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2} cos θ = 2 e i θ + e − i θ
sin θ = e i θ − e − i θ 2 i \sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i} sin θ = 2 i e i θ − e − i θ
Problem. Express ( 1 + i ) 5 ( 1 − i ) 3 \dfrac{(1 + i)^5}{(1 - i)^3} ( 1 − i ) 3 ( 1 + i ) 5 a + b i a + bi a + bi
1 + i = \sqrt{2}\,e^{i\pi/4}$$1 - i = \sqrt{2}\,e^{-i\pi/4} .
( 1 + i ) 5 ( 1 − i ) 3 = ( 2 ) 5 e i 5 π / 4 ( 2 ) 3 e − i 3 π / 4 = 4 2 e i 2 π = 4 2 \frac{(1 + i)^5}{(1 - i)^3} = \frac{(\sqrt{2})^5\,e^{i5\pi/4}}{(\sqrt{2})^3\,e^{-i3\pi/4}} = 4\sqrt{2}\,e^{i2\pi} = 4\sqrt{2} ( 1 − i ) 3 ( 1 + i ) 5 = ( 2 ) 3 e − i 3 π /4 ( 2 ) 5 e i 5 π /4 = 4 2 e i 2 π = 4 2
In Cartesian form: 4 2 + 0 i 4\sqrt{2} + 0i 4 2 + 0 i
Problem. Find all solutions to e z = 1 + i 3 e^z = 1 + i\sqrt{3} e z = 1 + i 3
1 + i 3 = 2 e i π / 3 1 + i\sqrt{3} = 2\,e^{i\pi/3} 1 + i 3 = 2 e iπ /3
So e x + i y = e x e i y = 2 e i ( π / 3 + 2 k π ) e^{x + iy} = e^x\,e^{iy} = 2\,e^{i(\pi/3 + 2k\pi)} e x + i y = e x e i y = 2 e i ( π /3 + 2 k π )
Equating moduli: e x = 2 ⟹ x = ln 2 e^x = 2 \implies x = \ln 2 e x = 2 ⟹ x = ln 2
Equating arguments: y = π 3 + 2 k π y = \dfrac{\pi}{3} + 2k\pi y = 3 π + 2 k π k ∈ Z k \in \mathbb{Z} k ∈ Z
z = ln 2 + i ( π 3 + 2 k π ) , k ∈ Z z = \ln 2 + i\!\left(\frac{\pi}{3} + 2k\pi\right), \quad k \in \mathbb{Z} z = ln 2 + i ( 3 π + 2 k π ) , k ∈ Z
The form $e^z = w$ have infinitely many solutions. Always include the general solution with $2k\pi$. Express z = − 3 + i z = -\sqrt{3} + i z = − 3 + i z 8 z^8 z 8
Solution r = 3 + 1 = 2 r = \sqrt{3 + 1} = 2 r = 3 + 1 = 2 θ = π − π 6 = 5 π 6 \theta = \pi - \dfrac{\pi}{6} = \dfrac{5\pi}{6} θ = π − 6 π = 6 5 π
z = 2 e i 5 π / 6 z = 2\,e^{i5\pi/6} z = 2 e i 5 π /6
z 8 = 2 8 e i 40 π / 6 = 256 e i 20 π / 3 = 256 e i ( 6 π + 2 π / 3 ) = 256 e i 2 π / 3 = 256 ( − 1 2 + i 3 2 ) = − 128 + 128 i 3 z^8 = 2^8\,e^{i40\pi/6} = 256\,e^{i20\pi/3} = 256\,e^{i(6\pi + 2\pi/3)} = 256\,e^{i2\pi/3} = 256\!\left(-\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = -128 + 128i\sqrt{3} z 8 = 2 8 e i 40 π /6 = 256 e i 20 π /3 = 256 e i ( 6 π + 2 π /3 ) = 256 e i 2 π /3 = 256 ( − 2 1 + i 2 3 ) = − 128 + 128 i 3
Find all fifth roots of 16 + 16 i 16 + 16i 16 + 16 i
Solution 16 + 16 i = 16 2 e i π / 4 16 + 16i = 16\sqrt{2}\,e^{i\pi/4} 16 + 16 i = 16 2 e iπ /4
z k = ( 16 2 ) 1 / 5 e i ( π / 4 + 2 k π ) / 5 = 2 1 / 2 + 3 / 5 e i ( π + 8 k π ) / 20 = 2 11 / 10 e i ( π + 8 k π ) / 20 z_k = (16\sqrt{2})^{1/5}\,e^{i(\pi/4 + 2k\pi)/5} = 2^{1/2 + 3/5}\,e^{i(\pi + 8k\pi)/20} = 2^{11/10}\,e^{i(\pi + 8k\pi)/20} z k = ( 16 2 ) 1/5 e i ( π /4 + 2 k π ) /5 = 2 1/2 + 3/5 e i ( π + 8 k π ) /20 = 2 11/10 e i ( π + 8 k π ) /20
For k = 0 , 1 , 2 , 3 , 4 k = 0, 1, 2, 3, 4 k = 0 , 1 , 2 , 3 , 4
The sum is 2 11 / 10 ∑ k = 0 4 e i ( π + 8 k π ) / 20 = 2 11 / 10 ⋅ e i π / 20 ∑ k = 0 4 ( e i 8 π / 20 ) k 2^{11/10}\displaystyle\sum_{k=0}^{4}e^{i(\pi + 8k\pi)/20} = 2^{11/10} \cdot e^{i\pi/20}\sum_{k=0}^{4}(e^{i8\pi/20})^k 2 11/10 k = 0 ∑ 4 e i ( π + 8 k π ) /20 = 2 11/10 ⋅ e iπ /20 k = 0 ∑ 4 ( e i 8 π /20 ) k
Since ω = e i 2 π / 5 \omega = e^{i2\pi/5} ω = e i 2 π /5 ω 5 = 1 \omega^5 = 1 ω 5 = 1 ω ≠ 1 \omega \neq 1 ω = 1
∑ k = 0 4 ω k = 1 − ω 5 1 − ω = 0 \sum_{k=0}^{4}\omega^k = \dfrac{1 - \omega^5}{1 - \omega} = 0 ∑ k = 0 4 ω k = 1 − ω 1 − ω 5 = 0
Find the Cartesian equation of the locus ∣ z − 1 + 2 i ∣ = ∣ z − 3 − 4 i ∣ |z - 1 + 2i| = |z - 3 - 4i| ∣ z − 1 + 2 i ∣ = ∣ z − 3 − 4 i ∣
Solution Let z = x + i y z = x + iy z = x + i y
( x − 1 ) 2 + ( y + 2 ) 2 = ( x − 3 ) 2 + ( y − 4 ) 2 (x - 1)^2 + (y + 2)^2 = (x - 3)^2 + (y - 4)^2 ( x − 1 ) 2 + ( y + 2 ) 2 = ( x − 3 ) 2 + ( y − 4 ) 2
x 2 − 2 x + 1 + y 2 + 4 y + 4 = x 2 − 6 x + 9 + y 2 − 8 y + 16 x^2 - 2x + 1 + y^2 + 4y + 4 = x^2 - 6x + 9 + y^2 - 8y + 16 x 2 − 2 x + 1 + y 2 + 4 y + 4 = x 2 − 6 x + 9 + y 2 − 8 y + 16
− 2 x + 4 y + 5 = − 6 x − 8 y + 25 -2x + 4y + 5 = -6x - 8y + 25 − 2 x + 4 y + 5 = − 6 x − 8 y + 25
4 x + 12 y = 20 ⟹ x + 3 y = 5 4x + 12y = 20 \implies x + 3y = 5 4 x + 12 y = 20 ⟹ x + 3 y = 5
This is a straight line (the perpendicular bisector of the segment joining 1 − 2 i 1 - 2i 1 − 2 i 3 + 4 i 3 + 4i 3 + 4 i
Use Euler’s formula to show that cos 4 θ = 3 + 4 cos 2 θ + cos 4 θ 8 \cos^4\theta = \dfrac{3 + 4\cos 2\theta + \cos 4\theta}{8} cos 4 θ = 8 3 + 4 cos 2 θ + cos 4 θ
Solution cos θ = e i θ + e − i θ 2 \cos\theta = \dfrac{e^{i\theta} + e^{-i\theta}}{2} cos θ = 2 e i θ + e − i θ cos 4 θ = 1 16 ( e i θ + e − i θ ) 4 \cos^4\theta = \dfrac{1}{16}(e^{i\theta} + e^{-i\theta})^4 cos 4 θ = 16 1 ( e i θ + e − i θ ) 4
= 1 16 ( e 4 i θ + 4 e 2 i θ + 6 + 4 e − 2 i θ + e − 4 i θ ) = \dfrac{1}{16}(e^{4i\theta} + 4e^{2i\theta} + 6 + 4e^{-2i\theta} + e^{-4i\theta}) = 16 1 ( e 4 i θ + 4 e 2 i θ + 6 + 4 e − 2 i θ + e − 4 i θ )
= 1 16 ( 2 cos 4 θ + 8 cos 2 θ + 6 ) = \dfrac{1}{16}(2\cos 4\theta + 8\cos 2\theta + 6) = 16 1 ( 2 cos 4 θ + 8 cos 2 θ + 6 )
= 3 + 4 cos 2 θ + cos 4 θ 8 = \dfrac{3 + 4\cos 2\theta + \cos 4\theta}{8} = 8 3 + 4 cos 2 θ + cos 4 θ
Theorem. The product of all n n n ( − 1 ) n − 1 (-1)^{n-1} ( − 1 ) n − 1
Proof. The n n n z n − 1 = 0 z^n - 1 = 0 z n − 1 = 0 n n n
∏ k = 0 n − 1 z k = ( − 1 ) n ⋅ − 1 1 = ( − 1 ) n − 1 ■ \prod_{k=0}^{n-1} z_k = (-1)^n \cdot \frac{-1}{1} = (-1)^{n-1} \quad \blacksquare ∏ k = 0 n − 1 z k = ( − 1 ) n ⋅ 1 − 1 = ( − 1 ) n − 1 ■
Theorem. If p ( z ) p(z) p ( z ) p ( α ) = 0 p(\alpha) = 0 p ( α ) = 0 p ( α ‾ ) = 0 p(\overline{\alpha}) = 0 p ( α ) = 0
Proof. Let p ( z ) = a n z n + ⋯ + a 1 z + a 0 p(z) = a_n z^n + \cdots + a_1 z + a_0 p ( z ) = a n z n + ⋯ + a 1 z + a 0 a i ∈ R a_i \in \mathbb{R} a i ∈ R
p ( α ‾ ) = a n α ‾ n + ⋯ + a 1 α ‾ + a 0 = a n ‾ α n ‾ + ⋯ + a 1 ‾ α ‾ + a 0 ‾ p(\overline{\alpha}) = a_n \overline{\alpha}^n + \cdots + a_1 \overline{\alpha} + a_0 = \overline{a_n}\,\overline{\alpha^n} + \cdots + \overline{a_1}\,\overline{\alpha} + \overline{a_0} p ( α ) = a n α n + ⋯ + a 1 α + a 0 = a n α n + ⋯ + a 1 α + a 0
= a n α n + ⋯ + a 1 α + a 0 ‾ = p ( α ) ‾ = 0 ‾ = 0 ■ = \overline{a_n \alpha^n + \cdots + a_1 \alpha + a_0} = \overline{p(\alpha)} = \overline{0} = 0 \quad \blacksquare = a n α n + ⋯ + a 1 α + a 0 = p ( α ) = 0 = 0 ■
Proof. Using the exponential form, let z 1 = r 1 e i θ 1 z_1 = r_1 e^{i\theta_1} z 1 = r 1 e i θ 1 z 2 = r 2 e i θ 2 z_2 = r_2 e^{i\theta_2} z 2 = r 2 e i θ 2
∣ z 1 + z 2 ∣ 2 = ( z 1 + z 2 ) ( z 1 + z 2 ) ‾ = ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 + z 1 z 2 ‾ + z 1 ‾ z 2 |z_1 + z_2|^2 = (z_1 + z_2)\overline{(z_1 + z_2)} = |z_1|^2 + |z_2|^2 + z_1\overline{z_2} + \overline{z_1}z_2 ∣ z 1 + z 2 ∣ 2 = ( z 1 + z 2 ) ( z 1 + z 2 ) = ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 + z 1 z 2 + z 1 z 2
= ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 + 2 R e ( z 1 z 2 ‾ ) ≤ ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 + 2 ∣ z 1 ∣ ∣ z 2 ∣ = ( ∣ z 1 ∣ + ∣ z 2 ∣ ) 2 = |z_1|^2 + |z_2|^2 + 2\,\mathrm{Re}(z_1\overline{z_2}) \leq |z_1|^2 + |z_2|^2 + 2|z_1||z_2| = (|z_1| + |z_2|)^2 = ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 + 2 Re ( z 1 z 2 ) ≤ ∣ z 1 ∣ 2 + ∣ z 2 ∣ 2 + 2∣ z 1 ∣∣ z 2 ∣ = ( ∣ z 1 ∣ + ∣ z 2 ∣ ) 2
Since R e ( w ) ≤ ∣ w ∣ \mathrm{Re}(w) \leq |w| Re ( w ) ≤ ∣ w ∣ w w w ■ \blacksquare ■
1. **Argument range:** Always specify whether your argument is the principal value $(-\pi, \pi]$ or the general value. When multiplying or dividing complex numbers, the resulting argument may fall outside the principal range and must be adjusted. 2. **Roots of negative numbers:** When finding roots of negative real numbers, the argument is $\pi$ (not $-\pi$), and the roots are distributed starting from angle $\pi/n$. 3. **Locus regions:** $|z - z_0| = r$ is a circle (boundary only). Use $\leq$ or $\geq$ for the interior or exterior including the boundary. Shade carefully in Argand diagrams. 4. **Forgetting all roots:** For $z^n = w$There are exactly $n$ distinct roots. Do not forget to include $k = 0, 1, \ldots, n-1$.The complex numbers z 1 z_1 z 1 z 2 z_2 z 2 z 1 = 1 + i z_1 = 1 + i z 1 = 1 + i z 2 = 1 − 3 i z_2 = 1 - \sqrt{3}i z 2 = 1 − 3 i
(a) Find z 1 z 2 z_1 z_2 z 1 z 2 r e i θ re^{i\theta} r e i θ − π < θ ≤ π -\pi < \theta \leq \pi − π < θ ≤ π
(b) Hence find z 1 4 z 2 2 \dfrac{z_1^4}{z_2^2} z 2 2 z 1 4
Solution (a) z 1 = 2 e i π / 4 z_1 = \sqrt{2}\,e^{i\pi/4} z 1 = 2 e iπ /4 z 2 = 2 e − i π / 3 z_2 = 2\,e^{-i\pi/3} z 2 = 2 e − iπ /3
z 1 z 2 = 2 2 e i ( π / 4 − π / 3 ) = 2 2 e − i π / 12 z_1 z_2 = 2\sqrt{2}\,e^{i(\pi/4 - \pi/3)} = 2\sqrt{2}\,e^{-i\pi/12} z 1 z 2 = 2 2 e i ( π /4 − π /3 ) = 2 2 e − iπ /12
(b) z 1 4 z 2 2 = ( 2 ) 4 e i π 2 2 e − i 2 π / 3 = 4 e i π 4 e − i 2 π / 3 = e i 5 π / 3 \dfrac{z_1^4}{z_2^2} = \dfrac{(\sqrt{2})^4\,e^{i\pi}}{2^2\,e^{-i2\pi/3}} = \dfrac{4\,e^{i\pi}}{4\,e^{-i2\pi/3}} = e^{i5\pi/3} z 2 2 z 1 4 = 2 2 e − i 2 π /3 ( 2 ) 4 e iπ = 4 e − i 2 π /3 4 e iπ = e i 5 π /3
In Cartesian form: e i 5 π / 3 = cos 5 π 3 + i sin 5 π 3 = 1 2 − i 3 2 e^{i5\pi/3} = \cos\dfrac{5\pi}{3} + i\sin\dfrac{5\pi}{3} = \dfrac{1}{2} - i\dfrac{\sqrt{3}}{2} e i 5 π /3 = cos 3 5 π + i sin 3 5 π = 2 1 − i 2 3
(a) Show that the roots of z 4 + 16 = 0 z^4 + 16 = 0 z 4 + 16 = 0 z = 2 e i ( π + 2 k π ) / 4 z = \sqrt{2}\,e^{i(\pi + 2k\pi)/4} z = 2 e i ( π + 2 k π ) /4 k = 0 , 1 , 2 , 3 k = 0, 1, 2, 3 k = 0 , 1 , 2 , 3
(b) Find the Cartesian equations of the perpendicular bisector of the line segment joining two Consecutive roots.
Solution (a) z 4 = − 16 = 16 e i π z^4 = -16 = 16\,e^{i\pi} z 4 = − 16 = 16 e iπ z = 16 1 / 4 e i ( π + 2 k π ) / 4 = 2 e i ( π + 2 k π ) / 4 z = 16^{1/4}\,e^{i(\pi + 2k\pi)/4} = 2\,e^{i(\pi + 2k\pi)/4} z = 1 6 1/4 e i ( π + 2 k π ) /4 = 2 e i ( π + 2 k π ) /4 k = 0 , 1 , 2 , 3 k = 0, 1, 2, 3 k = 0 , 1 , 2 , 3
Note: 2 \sqrt{2} 2 2 2 2 16 1 / 4 = 2 16^{1/4} = 2 1 6 1/4 = 2
The four roots are:
k = 0 k = 0 k = 0 2 e i π / 4 = 2 + i 2 2e^{i\pi/4} = \sqrt{2} + i\sqrt{2} 2 e iπ /4 = 2 + i 2 k = 1 k = 1 k = 1 2 e i 3 π / 4 = − 2 + i 2 2e^{i3\pi/4} = -\sqrt{2} + i\sqrt{2} 2 e i 3 π /4 = − 2 + i 2 k = 2 k = 2 k = 2 2 e i 5 π / 4 = − 2 − i 2 2e^{i5\pi/4} = -\sqrt{2} - i\sqrt{2} 2 e i 5 π /4 = − 2 − i 2 k = 3 k = 3 k = 3 2 e i 7 π / 4 = 2 − i 2 2e^{i7\pi/4} = \sqrt{2} - i\sqrt{2} 2 e i 7 π /4 = 2 − i 2 (b) Consecutive roots z 0 = 2 + i 2 z_0 = \sqrt{2} + i\sqrt{2} z 0 = 2 + i 2 z 1 = − 2 + i 2 z_1 = -\sqrt{2} + i\sqrt{2} z 1 = − 2 + i 2
The midpoint is ( 0 , 2 ) (0, \sqrt{2}) ( 0 , 2 ) x = 0 x = 0 x = 0
The point P P P z z z ∣ z − 4 − 3 i ∣ = 2 ∣ z − 1 + i ∣ |z - 4 - 3i| = 2|z - 1 + i| ∣ z − 4 − 3 i ∣ = 2∣ z − 1 + i ∣
(a) Show that the locus of P P P
(b) Find the minimum and maximum values of ∣ z ∣ |z| ∣ z ∣
Solution (a) Let z = x + i y z = x + iy z = x + i y
( x − 4 ) 2 + ( y − 3 ) 2 = 4 [ ( x − 1 ) 2 + ( y + 1 ) 2 ] (x - 4)^2 + (y - 3)^2 = 4\bigl[(x - 1)^2 + (y + 1)^2\bigr] ( x − 4 ) 2 + ( y − 3 ) 2 = 4 [ ( x − 1 ) 2 + ( y + 1 ) 2 ]
x 2 − 8 x + 16 + y 2 − 6 y + 9 = 4 x 2 − 8 x + 4 + 4 y 2 + 8 y + 4 x^2 - 8x + 16 + y^2 - 6y + 9 = 4x^2 - 8x + 4 + 4y^2 + 8y + 4 x 2 − 8 x + 16 + y 2 − 6 y + 9 = 4 x 2 − 8 x + 4 + 4 y 2 + 8 y + 4
3 x 2 + 3 y 2 + 12 y − 17 = 0 3x^2 + 3y^2 + 12y - 17 = 0 3 x 2 + 3 y 2 + 12 y − 17 = 0
x 2 + y 2 + 4 y = 17 3 x^2 + y^2 + 4y = \dfrac{17}{3} x 2 + y 2 + 4 y = 3 17
x 2 + ( y + 2 ) 2 = 17 3 + 4 = 29 3 x^2 + (y + 2)^2 = \dfrac{17}{3} + 4 = \dfrac{29}{3} x 2 + ( y + 2 ) 2 = 3 17 + 4 = 3 29
Centre: ( 0 , − 2 ) (0, -2) ( 0 , − 2 ) 29 / 3 \sqrt{29/3} 29/3
(b) ∣ z ∣ |z| ∣ z ∣
Distance from origin to centre: 0 2 + ( − 2 ) 2 = 2 \sqrt{0^2 + (-2)^2} = 2 0 2 + ( − 2 ) 2 = 2
Minimum ∣ z ∣ = 2 − 29 / 3 |z| = 2 - \sqrt{29/3} ∣ z ∣ = 2 − 29/3 29 / 3 ≈ 3.11 > 2 \sqrt{29/3} \approx 3.11 > 2 29/3 ≈ 3.11 > 2 ∣ radius − d ∣ = 29 / 3 − 2 |\text{radius} - d| = \sqrt{29/3} - 2 ∣ radius − d ∣ = 29/3 − 2
Maximum ∣ z ∣ = 2 + 29 / 3 |z| = 2 + \sqrt{29/3} ∣ z ∣ = 2 + 29/3
Problem. The sixth roots of unity are the roots of z 6 − 1 = 0 z^6 - 1 = 0 z 6 − 1 = 0 z 6 − 1 z^6 - 1 z 6 − 1 ( z − 1 ) ( z + 1 ) ( z 2 − z + 1 ) ( z 2 + z + 1 ) (z-1)(z+1)(z^2-z+1)(z^2+z+1) ( z − 1 ) ( z + 1 ) ( z 2 − z + 1 ) ( z 2 + z + 1 )
Solution. The sixth roots of unity are e i k π / 3 e^{ik\pi/3} e ik π /3 k = 0 , 1 , … , 5 k = 0, 1, \ldots, 5 k = 0 , 1 , … , 5
k = 0 k = 0 k = 0 z = 1 z = 1 z = 1 ( z − 1 ) (z - 1) ( z − 1 ) k = 3 k = 3 k = 3 z = e i π = − 1 z = e^{i\pi} = -1 z = e iπ = − 1 ( z + 1 ) (z + 1) ( z + 1 ) k = 1 , 5 k = 1, 5 k = 1 , 5 z = e i π / 3 z = e^{i\pi/3} z = e iπ /3 z = e i 5 π / 3 z = e^{i5\pi/3} z = e i 5 π /3 z 2 − 2 cos ( π / 3 ) z + 1 = z 2 − z + 1 z^2 - 2\cos(\pi/3)z + 1 = z^2 - z + 1 z 2 − 2 cos ( π /3 ) z + 1 = z 2 − z + 1 k = 2 , 4 k = 2, 4 k = 2 , 4 z = e i 2 π / 3 z = e^{i2\pi/3} z = e i 2 π /3 z = e i 4 π / 3 z = e^{i4\pi/3} z = e i 4 π /3 z 2 − 2 cos ( 2 π / 3 ) z + 1 = z 2 + z + 1 z^2 - 2\cos(2\pi/3)z + 1 = z^2 + z + 1 z 2 − 2 cos ( 2 π /3 ) z + 1 = z 2 + z + 1
Therefore z 6 − 1 = ( z − 1 ) ( z + 1 ) ( z 2 − z + 1 ) ( z 2 + z + 1 ) z^6 - 1 = (z-1)(z+1)(z^2 - z + 1)(z^2 + z + 1) z 6 − 1 = ( z − 1 ) ( z + 1 ) ( z 2 − z + 1 ) ( z 2 + z + 1 ) ■ \blacksquare ■
Problem. Use De Moivre’s theorem to express cos 5 θ \cos 5\theta cos 5 θ cos θ \cos\theta cos θ
Solution. ( cos θ + i sin θ ) 5 = cos 5 θ + i sin 5 θ (\cos\theta + i\sin\theta)^5 = \cos 5\theta + i\sin 5\theta ( cos θ + i sin θ ) 5 = cos 5 θ + i sin 5 θ
Expanding the LHS using the binomial theorem:
( cos θ + i sin θ ) 5 = cos 5 θ + 5 i cos 4 θ sin θ − 10 cos 3 θ sin 2 θ − 10 i cos 2 θ sin 3 θ + 5 cos θ sin 4 θ + i sin 5 θ (\cos\theta + i\sin\theta)^5 = \cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta ( cos θ + i sin θ ) 5 = cos 5 θ + 5 i cos 4 θ sin θ − 10 cos 3 θ sin 2 θ − 10 i cos 2 θ sin 3 θ + 5 cos θ sin 4 θ + i sin 5 θ
Equating real parts and using sin 2 θ = 1 − cos 2 θ \sin^2\theta = 1 - \cos^2\theta sin 2 θ = 1 − cos 2 θ
cos 5 θ = cos 5 θ − 10 cos 3 θ ( 1 − cos 2 θ ) + 5 cos θ ( 1 − cos 2 θ ) 2 \cos 5\theta = \cos^5\theta - 10\cos^3\theta(1-\cos^2\theta) + 5\cos\theta(1-\cos^2\theta)^2 cos 5 θ = cos 5 θ − 10 cos 3 θ ( 1 − cos 2 θ ) + 5 cos θ ( 1 − cos 2 θ ) 2
= cos 5 θ − 10 cos 3 θ + 10 cos 5 θ + 5 cos θ − 10 cos 3 θ + 5 cos 5 θ = \cos^5\theta - 10\cos^3\theta + 10\cos^5\theta + 5\cos\theta - 10\cos^3\theta + 5\cos^5\theta = cos 5 θ − 10 cos 3 θ + 10 cos 5 θ + 5 cos θ − 10 cos 3 θ + 5 cos 5 θ
= 16 cos 5 θ − 20 cos 3 θ + 5 cos θ = 16\cos^5\theta - 20\cos^3\theta + 5\cos\theta = 16 cos 5 θ − 20 cos 3 θ + 5 cos θ
Problem. On an Argand diagram, shade the region R R R ∣ z − 3 i ∣ ≤ 2 |z - 3i| \leq 2 ∣ z − 3 i ∣ ≤ 2 0 ≤ arg ( z − i ) ≤ π / 4 0 \leq \arg(z - i) \leq \pi/4 0 ≤ arg ( z − i ) ≤ π /4
Solution. ∣ z − 3 i ∣ ≤ 2 |z - 3i| \leq 2 ∣ z − 3 i ∣ ≤ 2 3 i 3i 3 i 2 2 2
arg ( z − i ) = α \arg(z - i) = \alpha arg ( z − i ) = α i i i α \alpha α
The constraint 0 ≤ arg ( z − i ) ≤ π / 4 0 \leq \arg(z - i) \leq \pi/4 0 ≤ arg ( z − i ) ≤ π /4 i i i 45 ∘ 45^\circ 4 5 ∘ i i i
The intersection of the disc with this sector gives R R R arg ( z − i ) = π / 4 \arg(z - i) = \pi/4 arg ( z − i ) = π /4 i i i ( 1 , 1 ) (1, 1) ( 1 , 1 ) y − 1 = x y - 1 = x y − 1 = x y = x + 1 y = x + 1 y = x + 1
The disc boundary ( x ) 2 + ( y − 3 ) 2 = 4 (x)^2 + (y - 3)^2 = 4 ( x ) 2 + ( y − 3 ) 2 = 4 y = x + 1 y = x + 1 y = x + 1
x 2 + ( x + 1 − 3 ) 2 = 4 ⟹ x 2 + ( x − 2 ) 2 = 4 ⟹ 2 x 2 − 4 x = 0 ⟹ x = 0 or x = 2 x^2 + (x + 1 - 3)^2 = 4 \implies x^2 + (x-2)^2 = 4 \implies 2x^2 - 4x = 0 \implies x = 0 \text{ or } x = 2 x 2 + ( x + 1 − 3 ) 2 = 4 ⟹ x 2 + ( x − 2 ) 2 = 4 ⟹ 2 x 2 − 4 x = 0 ⟹ x = 0 or x = 2
So the intersection points are ( 0 , 1 ) = i (0, 1) = i ( 0 , 1 ) = i ( 2 , 3 ) (2, 3) ( 2 , 3 )
Problem. The transformation T T T z z z w w w w = z + 1 z − 1 w = \dfrac{z + 1}{z - 1} w = z − 1 z + 1 R e ( z ) = 2 \mathrm{Re}(z) = 2 Re ( z ) = 2 T T T
Solution. Let z = 2 + i y z = 2 + iy z = 2 + i y
w = 2 + i y + 1 2 + i y − 1 = 3 + i y 1 + i y = ( 3 + i y ) ( 1 − i y ) 1 + y 2 = 3 + y 2 + i ( y − 3 y ) 1 + y 2 = 3 + y 2 − 2 i y 1 + y 2 w = \frac{2 + iy + 1}{2 + iy - 1} = \frac{3 + iy}{1 + iy} = \frac{(3 + iy)(1 - iy)}{1 + y^2} = \frac{3 + y^2 + i(y - 3y)}{1 + y^2} = \frac{3 + y^2 - 2iy}{1 + y^2} w = 2 + i y − 1 2 + i y + 1 = 1 + i y 3 + i y = 1 + y 2 ( 3 + i y ) ( 1 − i y ) = 1 + y 2 3 + y 2 + i ( y − 3 y ) = 1 + y 2 3 + y 2 − 2 i y
So u = 3 + y 2 1 + y 2 u = \dfrac{3 + y^2}{1 + y^2} u = 1 + y 2 3 + y 2 v = − 2 y 1 + y 2 v = \dfrac{-2y}{1 + y^2} v = 1 + y 2 − 2 y
Note that u = 1 + 2 1 + y 2 u = 1 + \dfrac{2}{1 + y^2} u = 1 + 1 + y 2 2 u ≥ 1 u \geq 1 u ≥ 1 1 + y 2 ≥ 1 1 + y^2 \geq 1 1 + y 2 ≥ 1
Also u − 1 = 2 1 + y 2 u - 1 = \dfrac{2}{1 + y^2} u − 1 = 1 + y 2 2 v 2 = 4 y 2 ( 1 + y 2 ) 2 = 4 y 2 ( 1 + y 2 ) 2 v^2 = \dfrac{4y^2}{(1+y^2)^2} = \dfrac{4y^2}{(1+y^2)^2} v 2 = ( 1 + y 2 ) 2 4 y 2 = ( 1 + y 2 ) 2 4 y 2
v 2 = ( u − 1 ) ⋅ 2 y 2 1 + y 2 v^2 = (u-1)\cdot\frac{2y^2}{1+y^2} v 2 = ( u − 1 ) ⋅ 1 + y 2 2 y 2
From v = − 2 y 1 + y 2 v = \dfrac{-2y}{1+y^2} v = 1 + y 2 − 2 y v 2 = 4 y 2 ( 1 + y 2 ) 2 v^2 = \dfrac{4y^2}{(1+y^2)^2} v 2 = ( 1 + y 2 ) 2 4 y 2
Since u − 1 = 2 1 + y 2 u - 1 = \dfrac{2}{1+y^2} u − 1 = 1 + y 2 2 ( u − 1 ) ( 1 + y 2 ) = 2 (u-1)(1+y^2) = 2 ( u − 1 ) ( 1 + y 2 ) = 2 1 + y 2 = 2 u − 1 1+y^2 = \dfrac{2}{u-1} 1 + y 2 = u − 1 2
v 2 = 4 y 2 ( 1 + y 2 ) 2 = 4 ( 2 u − 1 − 1 ) 4 ( u − 1 ) 2 = 8 − 2 ( u − 1 ) u − 1 4 ( u − 1 ) 2 = ( 10 − 2 u ) ( u − 1 ) 4 = ( 5 − u ) ( u − 1 ) 2 v^2 = \frac{4y^2}{(1+y^2)^2} = \frac{4\!\left(\frac{2}{u-1} - 1\right)}{\frac{4}{(u-1)^2}} = \frac{\frac{8 - 2(u-1)}{u-1}}{\frac{4}{(u-1)^2}} = \frac{(10 - 2u)(u-1)}{4} = \frac{(5-u)(u-1)}{2} v 2 = ( 1 + y 2 ) 2 4 y 2 = ( u − 1 ) 2 4 4 ( u − 1 2 − 1 ) = ( u − 1 ) 2 4 u − 1 8 − 2 ( u − 1 ) = 4 ( 10 − 2 u ) ( u − 1 ) = 2 ( 5 − u ) ( u − 1 )
The image is the arc of the circle defined by 2 v 2 = ( 5 − u ) ( u − 1 ) 2v^2 = (5-u)(u-1) 2 v 2 = ( 5 − u ) ( u − 1 ) u ≥ 1 u \geq 1 u ≥ 1 ( 3 , 0 ) (3, 0) ( 3 , 0 ) 2 2 2 w w w
Problem. Find all solutions to z 5 = 4 2 ( 1 + i ) z^5 = 4\sqrt{2}(1 + i) z 5 = 4 2 ( 1 + i )
Solution. 1 + i = 2 e i π / 4 1 + i = \sqrt{2}\,e^{i\pi/4} 1 + i = 2 e iπ /4 4 2 ( 1 + i ) = 8 e i π / 4 4\sqrt{2}(1 + i) = 8\,e^{i\pi/4} 4 2 ( 1 + i ) = 8 e iπ /4
z k = 8 1 / 5 e i ( π / 4 + 2 k π ) / 5 = 2 3 / 5 e i ( π + 8 k π ) / 20 , k = 0 , 1 , 2 , 3 , 4 z_k = 8^{1/5}\,e^{i(\pi/4 + 2k\pi)/5} = 2^{3/5}\,e^{i(\pi + 8k\pi)/20}, \quad k = 0, 1, 2, 3, 4 z k = 8 1/5 e i ( π /4 + 2 k π ) /5 = 2 3/5 e i ( π + 8 k π ) /20 , k = 0 , 1 , 2 , 3 , 4
The five roots lie on a circle of radius 2 3 / 5 2^{3/5} 2 3/5 π / 20 \pi/20 π /20
Problem. The cubic equation z 3 + a z 2 + b z + 12 = 0 z^3 + az^2 + bz + 12 = 0 z 3 + a z 2 + b z + 12 = 0 z = 1 + 2 i z = 1 + 2i z = 1 + 2 i a , b a, b a , b a a a b b b
Solution. Since a , b a, b a , b 1 − 2 i 1 - 2i 1 − 2 i
[ z − ( 1 + 2 i ) ] [ z − ( 1 − 2 i ) ] = ( z − 1 ) 2 + 4 = z 2 − 2 z + 5 [z - (1 + 2i)][z - (1 - 2i)] = (z-1)^2 + 4 = z^2 - 2z + 5 [ z − ( 1 + 2 i )] [ z − ( 1 − 2 i )] = ( z − 1 ) 2 + 4 = z 2 − 2 z + 5
Dividing z 3 + a z 2 + b z + 12 z^3 + az^2 + bz + 12 z 3 + a z 2 + b z + 12 z 2 − 2 z + 5 z^2 - 2z + 5 z 2 − 2 z + 5
z 3 + a z 2 + b z + 12 = ( z 2 − 2 z + 5 ) ( z − c ) z^3 + az^2 + bz + 12 = (z^2 - 2z + 5)(z - c) z 3 + a z 2 + b z + 12 = ( z 2 − 2 z + 5 ) ( z − c )
Expanding: z 3 − c z 2 − 2 z 2 + 2 c z + 5 z − 5 c = z 3 + ( − c − 2 ) z 2 + ( 2 c + 5 ) z − 5 c z^3 - cz^2 - 2z^2 + 2cz + 5z - 5c = z^3 + (-c-2)z^2 + (2c+5)z - 5c z 3 − c z 2 − 2 z 2 + 2 cz + 5 z − 5 c = z 3 + ( − c − 2 ) z 2 + ( 2 c + 5 ) z − 5 c
Comparing coefficients: -c - 2 = a$$2c + 5 = b$$-5c = 12 So c = − 12 / 5 c = -12/5 c = − 12/5
a = 12/5 - 2 = 2/5$$b = -24/5 + 5 = 1/5 .
The third root is z = c = − 12 / 5 z = c = -12/5 z = c = − 12/5
Check: ( 2 / 5 ) ( − 12 / 5 ) ( 12 / 5 ) ≠ − 12 (2/5)(-12/5)(12/5) \neq -12 ( 2/5 ) ( − 12/5 ) ( 12/5 ) = − 12 = ( 1 + 2 i ) ( 1 − 2 i ) ( − 12 / 5 ) = 5 × ( − 12 / 5 ) = − 12 = (1+2i)(1-2i)(-12/5) = 5 \times (-12/5) = -12 = ( 1 + 2 i ) ( 1 − 2 i ) ( − 12/5 ) = 5 × ( − 12/5 ) = − 12 = − 12 = -12 = − 12 = − ( − 12 ) / 1 = 12 = -(-12)/1 = 12 = − ( − 12 ) /1 = 12 − d / a = − 12 / 1 = − 12 -d/a = -12/1 = -12 − d / a = − 12/1 = − 12 12 12 12 z 3 + a z 2 + b z + 12 = 0 z^3 + az^2 + bz + 12 = 0 z 3 + a z 2 + b z + 12 = 0 d = 12 d = 12 d = 12 α β γ = − 12 / 1 = − 12 \alpha\beta\gamma = -12/1 = -12 α β γ = − 12/1 = − 12 5 × ( − 12 / 5 ) = − 12 5 \times (-12/5) = -12 5 × ( − 12/5 ) = − 12 ■ \blacksquare ■
Problem. The points A$$B$$C in the Argand diagram represent the complex numbers z_A = 2 + i$$z_B = 4 + 5i$$z_C = 6 + 2i . Show that triangle A B C ABC A B C
Solution. A B → = z B − z A = 2 + 4 i \overrightarrow{AB} = z_B - z_A = 2 + 4i A B = z B − z A = 2 + 4 i ∣ A B → ∣ = 4 + 16 = 2 5 |\overrightarrow{AB}| = \sqrt{4 + 16} = 2\sqrt{5} ∣ A B ∣ = 4 + 16 = 2 5
\overrightarrow{BC} = z_C - z_B = 2 - 3i$$|\overrightarrow{BC}| = \sqrt{4 + 9} = \sqrt{13} .
\overrightarrow{AC} = z_C - z_A = 4 + i$$|\overrightarrow{AC}| = \sqrt{16 + 1} = \sqrt{17} .
All three sides have different lengths, so the triangle is scalene , not isosceles. (If the Question intended different points, the same method applies.)
The area is 1 2 ∣ I m ( z A B ‾ ⋅ z A C ) ∣ \dfrac{1}{2}|\mathrm{Im}(\overline{z_{AB}} \cdot z_{AC})| 2 1 ∣ Im ( z A B ⋅ z A C ) ∣
z A B ‾ ⋅ z A C = ( 2 − 4 i ) ( 4 + i ) = 8 + 2 i − 16 i + 4 = 12 − 14 i \overline{z_{AB}} \cdot z_{AC} = (2 - 4i)(4 + i) = 8 + 2i - 16i + 4 = 12 - 14i z A B ⋅ z A C = ( 2 − 4 i ) ( 4 + i ) = 8 + 2 i − 16 i + 4 = 12 − 14 i
Area = 1 2 ∣ − 14 ∣ = 7 \text{Area} = \dfrac{1}{2}|-14| = 7 Area = 2 1 ∣ − 14∣ = 7
The matrix representation of complex numbers: a + b i ⟷ ( a − b b a ) a + bi \longleftrightarrow \begin{pmatrix}a & -b \\ b & a\end{pmatrix} a + bi ⟷ ( a b − b a ) Matrices and Transformations .
Modulus-argument form is equivalent to polar coordinates. The polar area formula applies to curves Described by ∣ z − z 0 ∣ = r ( θ ) |z - z_0| = r(\theta) ∣ z − z 0 ∣ = r ( θ ) Polar Coordinates .
Euler’s formula e i θ = cos θ + i sin θ e^{i\theta} = \cos\theta + i\sin\theta e i θ = cos θ + i sin θ Maclaurin and Taylor Series .
The n n n z n − 1 = ∏ k = 0 n − 1 ( z − ω k ) z^n - 1 = \prod_{k=0}^{n-1}(z - \omega^k) z n − 1 = ∏ k = 0 n − 1 ( z − ω k ) Further Algebra .
(a) Find all solutions of z 3 + 27 i = 0 z^3 + 27i = 0 z 3 + 27 i = 0
(b) Show that the sum of the three roots is zero.
Solution (a) z 3 = − 27 i = 27 e i ( 3 π / 2 + 2 k π ) z^3 = -27i = 27\,e^{i(3\pi/2 + 2k\pi)} z 3 = − 27 i = 27 e i ( 3 π /2 + 2 k π )
z k = 27 1 / 3 e i ( 3 π / 2 + 2 k π ) / 3 = 3 e i ( π / 2 + 2 k π / 3 ) , k = 0 , 1 , 2 z_k = 27^{1/3}\,e^{i(3\pi/2 + 2k\pi)/3} = 3\,e^{i(\pi/2 + 2k\pi/3)}, \quad k = 0, 1, 2 z k = 2 7 1/3 e i ( 3 π /2 + 2 k π ) /3 = 3 e i ( π /2 + 2 k π /3 ) , k = 0 , 1 , 2
k = 0 k = 0 k = 0 3 e i π / 2 = 3 i 3e^{i\pi/2} = 3i 3 e iπ /2 = 3 i k = 1 k = 1 k = 1 3 e i 7 π / 6 = 3 ( − 3 2 − i 2 ) 3e^{i7\pi/6} = 3\!\left(-\dfrac{\sqrt{3}}{2} - \dfrac{i}{2}\right) 3 e i 7 π /6 = 3 ( − 2 3 − 2 i ) k = 2 k = 2 k = 2 3 e i 11 π / 6 = 3 ( 3 2 − i 2 ) 3e^{i11\pi/6} = 3\!\left(\dfrac{\sqrt{3}}{2} - \dfrac{i}{2}\right) 3 e i 11 π /6 = 3 ( 2 3 − 2 i )
(b) Sum = 3 i + 3 ( − 3 2 − i 2 ) + 3 ( 3 2 − i 2 ) = 3 i − 3 i / 2 − 3 i / 2 = 0 = 3i + 3\!\left(-\dfrac{\sqrt{3}}{2} - \dfrac{i}{2}\right) + 3\!\left(\dfrac{\sqrt{3}}{2} - \dfrac{i}{2}\right) = 3i - 3i/2 - 3i/2 = 0 = 3 i + 3 ( − 2 3 − 2 i ) + 3 ( 2 3 − 2 i ) = 3 i − 3 i /2 − 3 i /2 = 0 ■ \blacksquare ■
The complex number w w w ∣ w + 2 i ∣ = 3 |w + 2i| = 3 ∣ w + 2 i ∣ = 3 arg w = π / 6 \arg w = \pi/6 arg w = π /6 w w w a + b i a + bi a + bi
Solution Let w = r e i π / 6 w = re^{i\pi/6} w = r e iπ /6 ∣ w + 2 i ∣ = 3 |w + 2i| = 3 ∣ w + 2 i ∣ = 3
∣ r e i π / 6 + 2 i ∣ = 3 |re^{i\pi/6} + 2i| = 3 ∣ r e iπ /6 + 2 i ∣ = 3
r cos ( π / 6 ) + i ( r sin ( π / 6 ) + 2 ) = r ( 3 2 ) + i ( r 2 + 2 ) r\cos(\pi/6) + i\!\left(r\sin(\pi/6) + 2\right) = r\!\left(\dfrac{\sqrt{3}}{2}\right) + i\!\left(\dfrac{r}{2} + 2\right) r cos ( π /6 ) + i ( r sin ( π /6 ) + 2 ) = r ( 2 3 ) + i ( 2 r + 2 )
3 r 2 4 + ( r 2 + 2 ) 2 = 9 \frac{3r^2}{4} + \left(\frac{r}{2} + 2\right)^2 = 9 4 3 r 2 + ( 2 r + 2 ) 2 = 9
3 r 2 4 + r 2 4 + 2 r + 4 = 9 ⟹ r 2 + 2 r − 5 = 0 \frac{3r^2}{4} + \frac{r^2}{4} + 2r + 4 = 9 \implies r^2 + 2r - 5 = 0 4 3 r 2 + 4 r 2 + 2 r + 4 = 9 ⟹ r 2 + 2 r − 5 = 0
r = − 2 ± 4 + 20 2 = − 1 ± 6 r = \frac{-2 \pm \sqrt{4 + 20}}{2} = -1 \pm \sqrt{6} r = 2 − 2 ± 4 + 20 = − 1 ± 6
Since r > 0 r > 0 r > 0 r = 6 − 1 r = \sqrt{6} - 1 r = 6 − 1
w = ( 6 − 1 ) e i π / 6 = ( 6 − 1 ) ( 3 2 + i 2 ) = ( 6 − 1 ) 3 2 + ( 6 − 1 ) i 2 w = (\sqrt{6} - 1)\,e^{i\pi/6} = (\sqrt{6} - 1)\!\left(\dfrac{\sqrt{3}}{2} + \dfrac{i}{2}\right) = \dfrac{(\sqrt{6}-1)\sqrt{3}}{2} + \dfrac{(\sqrt{6}-1)i}{2} w = ( 6 − 1 ) e iπ /6 = ( 6 − 1 ) ( 2 3 + 2 i ) = 2 ( 6 − 1 ) 3 + 2 ( 6 − 1 ) i
Use De Moivre’s theorem to find cos ( π / 12 ) \cos(\pi/12) cos ( π /12 ) sin ( π / 12 ) \sin(\pi/12) sin ( π /12 )
Solution cos ( π / 12 ) + i sin ( π / 12 ) = cos ( π / 6 ) + i sin ( π / 6 ) \cos(\pi/12) + i\sin(\pi/12) = \sqrt{\cos(\pi/6) + i\sin(\pi/6)} cos ( π /12 ) + i sin ( π /12 ) = cos ( π /6 ) + i sin ( π /6 )
cos ( π / 6 ) = 3 2 \cos(\pi/6) = \dfrac{\sqrt{3}}{2} cos ( π /6 ) = 2 3 sin ( π / 6 ) = 1 2 \sin(\pi/6) = \dfrac{1}{2} sin ( π /6 ) = 2 1
Let cos ( π / 12 ) + i sin ( π / 12 ) = a + b i \cos(\pi/12) + i\sin(\pi/12) = a + bi cos ( π /12 ) + i sin ( π /12 ) = a + bi a > 0 a > 0 a > 0 ( a + b i ) 2 = 3 2 + i 2 (a + bi)^2 = \dfrac{\sqrt{3}}{2} + \dfrac{i}{2} ( a + bi ) 2 = 2 3 + 2 i
a 2 − b 2 = 3 2 a^2 - b^2 = \dfrac{\sqrt{3}}{2} a 2 − b 2 = 2 3 2 a b = 1 2 2ab = \dfrac{1}{2} 2 ab = 2 1 b = 1 4 a b = \dfrac{1}{4a} b = 4 a 1
a 2 − 1 16 a 2 = 3 2 a^2 - \dfrac{1}{16a^2} = \dfrac{\sqrt{3}}{2} a 2 − 16 a 2 1 = 2 3
Multiplying by 16 a 2 16a^2 16 a 2 16 a 4 − 8 3 a 2 − 1 = 0 16a^4 - 8\sqrt{3}\,a^2 - 1 = 0 16 a 4 − 8 3 a 2 − 1 = 0
a 2 = 8 3 ± 192 + 64 32 = 8 3 ± 4 13 32 = 2 3 ± 13 8 a^2 = \dfrac{8\sqrt{3} \pm \sqrt{192 + 64}}{32} = \dfrac{8\sqrt{3} \pm 4\sqrt{13}}{32} = \dfrac{2\sqrt{3} \pm \sqrt{13}}{8} a 2 = 32 8 3 ± 192 + 64 = 32 8 3 ± 4 13 = 8 2 3 ± 13
Since a 2 ≤ 1 a^2 \leq 1 a 2 ≤ 1 a 2 = 2 3 + 13 8 a^2 = \dfrac{2\sqrt{3} + \sqrt{13}}{8} a 2 = 8 2 3 + 13
cos π 12 = 2 3 + 13 8 = 2 3 + 13 2 2 \cos\frac{\pi}{12} = \sqrt{\frac{2\sqrt{3} + \sqrt{13}}{8}} = \frac{\sqrt{2\sqrt{3} + \sqrt{13}}}{2\sqrt{2}} cos 12 π = 8 2 3 + 13 = 2 2 2 3 + 13
sin π 12 = − 2 3 + 13 8 = − 2 3 + 13 2 2 \sin\frac{\pi}{12} = \sqrt{\frac{-2\sqrt{3} + \sqrt{13}}{8}} = \frac{\sqrt{-2\sqrt{3} + \sqrt{13}}}{2\sqrt{2}} sin 12 π = 8 − 2 3 + 13 = 2 2 − 2 3 + 13
Prove by induction that for any positive integer n n n
∑ k = 0 n − 1 e i ( 2 k + 1 ) π / n = 0 \sum_{k=0}^{n-1} e^{i(2k+1)\pi/n} = 0 ∑ k = 0 n − 1 e i ( 2 k + 1 ) π / n = 0
Solution This is the sum of the n n n − 1 -1 − 1
Base case (n = 1 n = 1 n = 1 Sum = e i π = − 1 ≠ 0 = e^{i\pi} = -1 \neq 0 = e iπ = − 1 = 0
Let us reconsider: ∑ k = 0 n − 1 e i ( 2 k + 1 ) π / n \displaystyle\sum_{k=0}^{n-1}e^{i(2k+1)\pi/n} k = 0 ∑ n − 1 e i ( 2 k + 1 ) π / n e i π / n e^{i\pi/n} e iπ / n e i 2 π / n e^{i2\pi/n} e i 2 π / n
∑ k = 0 n − 1 e i ( 2 k + 1 ) π / n = e i π / n ⋅ 1 − e i 2 π 1 − e i 2 π / n = e i π / n ⋅ 1 − 1 1 − e i 2 π / n = 0 ■ \sum_{k=0}^{n-1}e^{i(2k+1)\pi/n} = e^{i\pi/n}\cdot\frac{1 - e^{i2\pi}}{1 - e^{i2\pi/n}} = e^{i\pi/n}\cdot\frac{1 - 1}{1 - e^{i2\pi/n}} = 0 \quad \blacksquare ∑ k = 0 n − 1 e i ( 2 k + 1 ) π / n = e iπ / n ⋅ 1 − e i 2 π / n 1 − e i 2 π = e iπ / n ⋅ 1 − e i 2 π / n 1 − 1 = 0 ■
(Since e i 2 π = 1 e^{i2\pi} = 1 e i 2 π = 1
Given z 1 = 2 + 3 i z_1 = 2 + 3i z 1 = 2 + 3 i z 2 = 1 − i z_2 = 1 - i z 2 = 1 − i z z z z_1$$z$$z_2 form An equilateral triangle, giving both possible values of z z z
Solution If z_1$$z$$z_2 form an equilateral triangle, then z z z z 2 − z 1 z_2 - z_1 z 2 − z 1 ± π / 3 \pm\pi/3 ± π /3 z 1 z_1 z 1
z = z 1 + ( z 2 − z 1 ) e ± i π / 3 z = z_1 + (z_2 - z_1)\,e^{\pm i\pi/3} z = z 1 + ( z 2 − z 1 ) e ± iπ /3
z 2 − z 1 = ( 1 − 2 ) + ( − 1 − 3 ) i = − 1 − 4 i z_2 - z_1 = (1 - 2) + (-1 - 3)i = -1 - 4i z 2 − z 1 = ( 1 − 2 ) + ( − 1 − 3 ) i = − 1 − 4 i
e i π / 3 = 1 2 + i 3 2 e^{i\pi/3} = \dfrac{1}{2} + i\dfrac{\sqrt{3}}{2} e iπ /3 = 2 1 + i 2 3
( − 1 − 4 i ) ( 1 2 + i 3 2 ) = − 1 + 4 3 2 + i − 4 − 3 2 (-1 - 4i)\!\left(\frac{1}{2} + i\frac{\sqrt{3}}{2}\right) = \frac{-1 + 4\sqrt{3}}{2} + i\frac{-4 - \sqrt{3}}{2} ( − 1 − 4 i ) ( 2 1 + i 2 3 ) = 2 − 1 + 4 3 + i 2 − 4 − 3
z = ( 2 + 3 i ) + − 1 + 4 3 2 + i − 4 − 3 2 = 3 + 4 3 2 + i 2 − 3 2 z = (2 + 3i) + \frac{-1 + 4\sqrt{3}}{2} + i\frac{-4 - \sqrt{3}}{2} = \frac{3 + 4\sqrt{3}}{2} + i\frac{2 - \sqrt{3}}{2} z = ( 2 + 3 i ) + 2 − 1 + 4 3 + i 2 − 4 − 3 = 2 3 + 4 3 + i 2 2 − 3
For the other orientation, e − i π / 3 = 1 2 − i 3 2 e^{-i\pi/3} = \dfrac{1}{2} - i\dfrac{\sqrt{3}}{2} e − iπ /3 = 2 1 − i 2 3
z = ( 2 + 3 i ) + − 1 − 4 3 2 + i − 4 + 3 2 = 3 − 4 3 2 + i 2 + 3 2 z = (2 + 3i) + \frac{-1 - 4\sqrt{3}}{2} + i\frac{-4 + \sqrt{3}}{2} = \frac{3 - 4\sqrt{3}}{2} + i\frac{2 + \sqrt{3}}{2} z = ( 2 + 3 i ) + 2 − 1 − 4 3 + i 2 − 4 + 3 = 2 3 − 4 3 + i 2 2 + 3
Problem. Express sin 4 θ \sin 4\theta sin 4 θ sin θ \sin\theta sin θ cos θ \cos\theta cos θ
Solution. ( cos θ + i sin θ ) 4 = cos 4 θ + i sin 4 θ (\cos\theta+i\sin\theta)^4 = \cos 4\theta + i\sin 4\theta ( cos θ + i sin θ ) 4 = cos 4 θ + i sin 4 θ
Imaginary parts: sin 4 θ = ( 4 1 ) cos 3 θ sin θ − ( 4 3 ) cos θ sin 3 θ \sin 4\theta = \binom{4}{1}\cos^3\theta\sin\theta - \binom{4}{3}\cos\theta\sin^3\theta sin 4 θ = ( 1 4 ) cos 3 θ sin θ − ( 3 4 ) cos θ sin 3 θ
sin 4 θ = 4 cos 3 θ sin θ − 4 cos θ sin 3 θ \boxed{\sin 4\theta = 4\cos^3\theta\sin\theta - 4\cos\theta\sin^3\theta} sin 4 θ = 4 cos 3 θ sin θ − 4 cos θ sin 3 θ
Problem. The 5th roots of unity are ω k \omega^k ω k k = 0 , … , 4 k = 0, \ldots, 4 k = 0 , … , 4 ω = e 2 π i / 5 \omega = e^{2\pi i/5} ω = e 2 π i /5
Solution. ω k = e 2 π i k / 5 \omega^k = e^{2\pi ik/5} ω k = e 2 π ik /5 0 , 72 ° , 144 ° , 216 ° , 288 ° 0, 72°, 144°, 216°, 288° 0 , 72° , 144° , 216° , 288°
Sum: ∑ k = 0 4 ω k = 1 − ω 5 1 − ω = 0 \displaystyle\sum_{k=0}^{4} \omega^k = \frac{1-\omega^5}{1-\omega} = 0 k = 0 ∑ 4 ω k = 1 − ω 1 − ω 5 = 0
Problem. Find the Cartesian equation of ∣ z − 1 + 2 i ∣ = 3 |z - 1 + 2i| = 3 ∣ z − 1 + 2 i ∣ = 3
Solution. Let z = x + i y z = x + iy z = x + i y
( x − 1 ) 2 + ( y + 2 ) 2 = 9 (x-1)^2 + (y+2)^2 = 9 ( x − 1 ) 2 + ( y + 2 ) 2 = 9
This is a circle with centre ( 1 , − 2 ) (1, -2) ( 1 , − 2 ) 3 3 3
Problem. z z z ∣ z − 3 i ∣ = 2 |z-3i| = 2 ∣ z − 3 i ∣ = 2 ∣ z ∣ |z| ∣ z ∣
Solution. The locus is a circle centred at 3 i 3i 3 i 2 2 2
∣ 3 i ∣ − 2 = 3 − 2 = 1 |3i| - 2 = 3 - 2 = \boxed{1} ∣3 i ∣ − 2 = 3 − 2 = 1
The point is z = i z = i z = i
Problem. The cubic x 3 + 3 x 2 + 7 x + 5 = 0 x^3 + 3x^2 + 7x + 5 = 0 x 3 + 3 x 2 + 7 x + 5 = 0
Solution. Trying x = − 1 x = -1 x = − 1 − 1 + 3 − 7 + 5 = 0 -1+3-7+5 = 0 − 1 + 3 − 7 + 5 = 0 x + 1 x+1 x + 1
x 3 + 3 x 2 + 7 x + 5 = ( x + 1 ) ( x 2 + 2 x + 5 ) x^3+3x^2+7x+5 = (x+1)(x^2+2x+5) x 3 + 3 x 2 + 7 x + 5 = ( x + 1 ) ( x 2 + 2 x + 5 )
x 2 + 2 x + 5 = 0 x^2+2x+5 = 0 x 2 + 2 x + 5 = 0 x = − 2 ± 4 − 20 2 = − 1 ± 2 i x = \dfrac{-2\pm\sqrt{4-20}}{2} = -1 \pm 2i x = 2 − 2 ± 4 − 20 = − 1 ± 2 i
x = − 1 , x = − 1 + 2 i , x = − 1 − 2 i \boxed{x = -1,\; x = -1+2i,\; x = -1-2i} x = − 1 , x = − 1 + 2 i , x = − 1 − 2 i
Problem. Under the transformation w = 1 z w = \dfrac{1}{z} w = z 1 x = 2 x = 2 x = 2
Solution. z = 2 + iy$$w = u + iv = \dfrac{1}{2+iy} = \dfrac{2-iy}{4+y^2} .
u = \dfrac{2}{4+y^2}$$v = \dfrac{-y}{4+y^2} .
Note: u 2 + v 2 = 4 + y 2 ( 4 + y 2 ) 2 = 1 4 + y 2 = u 2 u^2 + v^2 = \dfrac{4+y^2}{(4+y^2)^2} = \dfrac{1}{4+y^2} = \dfrac{u}{2} u 2 + v 2 = ( 4 + y 2 ) 2 4 + y 2 = 4 + y 2 1 = 2 u
So u 2 + v 2 = u 2 ⟹ ( u − 1 4 ) 2 + v 2 = 1 16 u^2 + v^2 = \dfrac{u}{2} \implies \left(u-\dfrac{1}{4}\right)^2 + v^2 = \dfrac{1}{16} u 2 + v 2 = 2 u ⟹ ( u − 4 1 ) 2 + v 2 = 16 1
The image is a circle with centre ( 1 4 , 0 ) \left(\dfrac{1}{4}, 0\right) ( 4 1 , 0 ) 1 4 \dfrac{1}{4} 4 1
Express cos 3 θ + cos θ \cos 3\theta + \cos\theta cos 3 θ + cos θ
Solution Using the sum-to-product formula: cos A + cos B = 2 cos ( A + B 2 ) cos ( A − B 2 ) \cos A + \cos B = 2\cos\!\left(\dfrac{A+B}{2}\right)\cos\!\left(\dfrac{A-B}{2}\right) cos A + cos B = 2 cos ( 2 A + B ) cos ( 2 A − B )
cos 3 θ + cos θ = 2 cos 2 θ cos θ = 2 cos θ cos 2 θ \cos 3\theta + \cos\theta = 2\cos 2\theta \cos\theta = \boxed{2\cos\theta\cos 2\theta} cos 3 θ + cos θ = 2 cos 2 θ cos θ = 2 cos θ cos 2 θ
Prove that the roots of z n = 1 z^n = 1 z n = 1
Solution z k = e 2 π i k / n z_k = e^{2\pi ik/n} z k = e 2 π ik / n k = 0 , 1 , … , n − 1 k = 0, 1, \ldots, n-1 k = 0 , 1 , … , n − 1
∣ z k ∣ = 1 |z_k| = 1 ∣ z k ∣ = 1 k k k
The angular separation between consecutive roots is 2 π n \dfrac{2\pi}{n} n 2 π
Therefore the roots are the vertices of a regular n n n ■ \blacksquare ■
Find the locus of points satisfying ∣ z − 1 ∣ = 2 ∣ z + 1 ∣ |z-1| = 2|z+1| ∣ z − 1∣ = 2∣ z + 1∣
Solution Let z = x + i y z = x+iy z = x + i y ( x − 1 ) 2 + y 2 = 4 [ ( x + 1 ) 2 + y 2 ] (x-1)^2+y^2 = 4[(x+1)^2+y^2] ( x − 1 ) 2 + y 2 = 4 [( x + 1 ) 2 + y 2 ]
x 2 − 2 x + 1 + y 2 = 4 x 2 + 8 x + 4 + 4 y 2 x^2-2x+1+y^2 = 4x^2+8x+4+4y^2 x 2 − 2 x + 1 + y 2 = 4 x 2 + 8 x + 4 + 4 y 2
0 = 3 x 2 + 10 x + 3 y 2 + 3 0 = 3x^2+10x+3y^2+3 0 = 3 x 2 + 10 x + 3 y 2 + 3
3 ( x 2 + 10 3 x ) + 3 y 2 = − 3 3\!\left(x^2+\dfrac{10}{3}x\right)+3y^2 = -3 3 ( x 2 + 3 10 x ) + 3 y 2 = − 3
3 ( x + 5 3 ) 2 + 3 y 2 = 25 3 − 3 = 16 3 3\!\left(x+\dfrac{5}{3}\right)^2 + 3y^2 = \dfrac{25}{3}-3 = \dfrac{16}{3} 3 ( x + 3 5 ) 2 + 3 y 2 = 3 25 − 3 = 3 16
( x + 5 3 ) 2 + y 2 = 16 9 \left(x+\dfrac{5}{3}\right)^2 + y^2 = \dfrac{16}{9} ( x + 3 5 ) 2 + y 2 = 9 16
A circle with centre ( − 5 3 , 0 ) \left(-\dfrac{5}{3}, 0\right) ( − 3 5 , 0 ) 4 3 \dfrac{4}{3} 3 4
The complex logarithm is multi-valued: ln z = ln ∣ z ∣ + i ( arg z + 2 k π ) \ln z = \ln|z| + i(\arg z + 2k\pi) ln z = ln ∣ z ∣ + i ( arg z + 2 k π ) k ∈ Z k \in \mathbb{Z} k ∈ Z
The principal value uses arg z ∈ ( − π , π ] \arg z \in (-\pi, \pi] arg z ∈ ( − π , π ]
z w = e w ln z z^w = e^{w\ln z} z w = e w l n z ln z \ln z ln z
Example: i i = e i ln i = e i ( i π / 2 ) = e − π / 2 ≈ 0.2079 i^i = e^{i\ln i} = e^{i(i\pi/2)} = e^{-\pi/2} \approx 0.2079 i i = e i l n i = e i ( iπ /2 ) = e − π /2 ≈ 0.2079
A Möbius transformation is w = a z + b c z + d w = \dfrac{az+b}{cz+d} w = cz + d a z + b a d − b c ≠ 0 ad-bc \neq 0 a d − b c = 0
Properties:
Maps circles and lines to circles and lines Preserves angles (conformal) Compositions of Möbius transformations are Möbius transformations Using e i θ = cos θ + i sin θ e^{i\theta} = \cos\theta + i\sin\theta e i θ = cos θ + i sin θ
cos n θ = ( e i θ + e − i θ 2 ) n \cos^n\theta = \left(\dfrac{e^{i\theta}+e^{-i\theta}}{2}\right)^n cos n θ = ( 2 e i θ + e − i θ ) n sin n θ = ( e i θ − e − i θ 2 i ) n \sin^n\theta = \left(\dfrac{e^{i\theta}-e^{-i\theta}}{2i}\right)^n sin n θ = ( 2 i e i θ − e − i θ ) n
These can be expanded to express cos n θ \cos^n\theta cos n θ sin n θ \sin^n\theta sin n θ
Express cos 4 θ \cos^4\theta cos 4 θ cos 2 θ \cos 2\theta cos 2 θ cos 4 θ \cos 4\theta cos 4 θ
Solution cos 4 θ = ( e i θ + e − i θ 2 ) 4 = e 4 i θ + 4 e 2 i θ + 6 + 4 e − 2 i θ + e − 4 i θ 16 \cos^4\theta = \left(\dfrac{e^{i\theta}+e^{-i\theta}}{2}\right)^4 = \dfrac{e^{4i\theta}+4e^{2i\theta}+6+4e^{-2i\theta}+e^{-4i\theta}}{16} cos 4 θ = ( 2 e i θ + e − i θ ) 4 = 16 e 4 i θ + 4 e 2 i θ + 6 + 4 e − 2 i θ + e − 4 i θ
= 2 cos 4 θ + 8 cos 2 θ + 6 16 = 3 8 + 1 2 cos 2 θ + 1 8 cos 4 θ = \dfrac{2\cos 4\theta + 8\cos 2\theta + 6}{16} = \dfrac{3}{8} + \dfrac{1}{2}\cos 2\theta + \dfrac{1}{8}\cos 4\theta = 16 2 cos 4 θ + 8 cos 2 θ + 6 = 8 3 + 2 1 cos 2 θ + 8 1 cos 4 θ
cos 4 θ = 3 8 + 1 2 cos 2 θ + 1 8 cos 4 θ \boxed{\cos^4\theta = \dfrac{3}{8} + \dfrac{1}{2}\cos 2\theta + \dfrac{1}{8}\cos 4\theta} cos 4 θ = 8 3 + 2 1 cos 2 θ + 8 1 cos 4 θ
Find the image of the line R e ( z ) = 2 \mathrm{Re}(z) = 2 Re ( z ) = 2 w = z − 1 z + 1 w = \dfrac{z-1}{z+1} w = z + 1 z − 1
Solution Let z = 2 + i y z = 2 + iy z = 2 + i y
w = 1 + i y 3 + i y = ( 1 + i y ) ( 3 − i y ) 9 + y 2 = 3 + y 2 + i ( 3 y − y ) 9 + y 2 = 3 + y 2 9 + y 2 + i 2 y 9 + y 2 w = \dfrac{1+iy}{3+iy} = \dfrac{(1+iy)(3-iy)}{9+y^2} = \dfrac{3+y^2+i(3y-y)}{9+y^2} = \dfrac{3+y^2}{9+y^2} + i\dfrac{2y}{9+y^2} w = 3 + i y 1 + i y = 9 + y 2 ( 1 + i y ) ( 3 − i y ) = 9 + y 2 3 + y 2 + i ( 3 y − y ) = 9 + y 2 3 + y 2 + i 9 + y 2 2 y
Let w = u + i v w = u+iv w = u + i v u = \dfrac{3+y^2}{9+y^2}$$v = \dfrac{2y}{9+y^2} .
Eliminating y y y u = 1 − 6 9 + y 2 u = 1 - \dfrac{6}{9+y^2} u = 1 − 9 + y 2 6 9 + y 2 = 6 1 − u 9+y^2 = \dfrac{6}{1-u} 9 + y 2 = 1 − u 6
v 2 = 4 y 2 ( 9 + y 2 ) 2 v^2 = \dfrac{4y^2}{(9+y^2)^2} v 2 = ( 9 + y 2 ) 2 4 y 2 y 2 = 6 1 − u − 9 = 6 − 9 + 9 u 1 − u = 9 u − 3 1 − u y^2 = \dfrac{6}{1-u}-9 = \dfrac{6-9+9u}{1-u} = \dfrac{9u-3}{1-u} y 2 = 1 − u 6 − 9 = 1 − u 6 − 9 + 9 u = 1 − u 9 u − 3
This is a circle (after simplification).
The n n n
They form a cyclic group under multiplication The sum of all n n n ∑ k = 0 n − 1 e 2 π i k / n = 0 \displaystyle\sum_{k=0}^{n-1} e^{2\pi ik/n} = 0 k = 0 ∑ n − 1 e 2 π ik / n = 0 Products of roots of unity are also roots of unity ln z = ln ∣ z ∣ + i arg z \ln z = \ln|z| + i\arg z ln z = ln ∣ z ∣ + i arg z arg z \arg z arg z
ln z = ln ∣ z ∣ + i ( θ + 2 k π ) \ln z = \ln|z| + i(\theta + 2k\pi) ln z = ln ∣ z ∣ + i ( θ + 2 k π ) k ∈ Z k \in \mathbb{Z} k ∈ Z
The principal value uses θ ∈ ( − π , π ] \theta \in (-\pi, \pi] θ ∈ ( − π , π ]
e i θ + e − i θ = 2 cos θ e^{i\theta} + e^{-i\theta} = 2\cos\theta e i θ + e − i θ = 2 cos θ
e i θ − e − i θ = 2 i sin θ e^{i\theta} - e^{-i\theta} = 2i\sin\theta e i θ − e − i θ = 2 i sin θ
cos θ = e i θ + e − i θ 2 \cos\theta = \dfrac{e^{i\theta}+e^{-i\theta}}{2} cos θ = 2 e i θ + e − i θ sin θ = e i θ − e − i θ 2 i \sin\theta = \dfrac{e^{i\theta}-e^{-i\theta}}{2i} sin θ = 2 i e i θ − e − i θ
Prove that ∑ k = 0 n − 1 e 2 π i k / n = 0 \displaystyle\sum_{k=0}^{n-1} e^{2\pi ik/n} = 0 k = 0 ∑ n − 1 e 2 π ik / n = 0 n ≥ 2 n \geq 2 n ≥ 2
Solution This is a geometric series with ratio r = e 2 π i / n ≠ 1 r = e^{2\pi i/n} \neq 1 r = e 2 π i / n = 1
∑ k = 0 n − 1 r k = r n − 1 r − 1 = e 2 π i − 1 e 2 π i / n − 1 = 1 − 1 e 2 π i / n − 1 = 0 \displaystyle\sum_{k=0}^{n-1} r^k = \frac{r^n - 1}{r - 1} = \frac{e^{2\pi i} - 1}{e^{2\pi i/n} - 1} = \frac{1 - 1}{e^{2\pi i/n} - 1} = 0 k = 0 ∑ n − 1 r k = r − 1 r n − 1 = e 2 π i / n − 1 e 2 π i − 1 = e 2 π i / n − 1 1 − 1 = 0 ■ \blacksquare ■
This topic explores fundamental concepts that shape our understanding of the world.
This topic covers the mathematical techniques and concepts related to complex numbers (extended), including key theorems, methods, and problem-solving approaches.
Key concepts include:
quadratic equations and the discriminant simultaneous equations polynomial division and the factor theorem partial fractions binomial expansion Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.
