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Complex Numbers -- Diagnostic Tests

Intuition

Section titled “Intuition”

This topic explores fundamental concepts that shape our understanding of the world.

Complex Numbers — Diagnostic Tests

Section titled “Complex Numbers — Diagnostic Tests”

Unit Tests

Section titled “Unit Tests”

UT-1: Modulus, Argument, and Argand Diagrams

Section titled “UT-1: Modulus, Argument, and Argand Diagrams”

Question: (a) Plot z1=3+4iz_1 = 3 + 4i and z2=−2+iz_2 = -2 + i on an Argand diagram. (b) Calculate |z_1|$$\arg(z_1)$$|z_2|And arg⁡(z2)\arg(z_2). (c) Express z1z_1 and z2z_2 in modulus-argument form r(cos⁡θ+isin⁡θ)r(\cos\theta + i\sin\theta). (d) Calculate z1z2z_1 z_2 and z1/z2z_1/z_2 using modulus-argument form.

Solution:

(b) ∣z1∣=9+16=5|z_1| = \sqrt{9 + 16} = 5. arg⁡(z1)=arctan⁡(4/3)≈0.927\arg(z_1) = \arctan(4/3) \approx 0.927 rad =53.1∘= 53.1^\circ. ∣z2∣=4+1=5|z_2| = \sqrt{4 + 1} = \sqrt{5}. arg⁡(z2)=π−arctan⁡(1/2)≈π−0.464=2.678\arg(z_2) = \pi - \arctan(1/2) \approx \pi - 0.464 = 2.678 rad =153.4∘= 153.4^\circ.

(c) z1=5(cos⁡0.927+isin⁡0.927)z_1 = 5(\cos 0.927 + i\sin 0.927). z2=5(cos⁡2.678+isin⁡2.678)z_2 = \sqrt{5}(\cos 2.678 + i\sin 2.678).

(d) z1z2z_1 z_2: ∣z1z2∣=55|z_1 z_2| = 5\sqrt{5}. arg⁡(z1z2)=0.927+2.678=3.605\arg(z_1 z_2) = 0.927 + 2.678 = 3.605 rad. z1z2=55(cos⁡3.605+isin⁡3.605)z_1 z_2 = 5\sqrt{5}(\cos 3.605 + i\sin 3.605).

z1/z2z_1/z_2: ∣z1/z2∣=5/5=5|z_1/z_2| = 5/\sqrt{5} = \sqrt{5}. arg⁡(z1/z2)=0.927−2.678=−1.751\arg(z_1/z_2) = 0.927 - 2.678 = -1.751 rad. z1/z2=5(cos⁡(−1.751)+isin⁡(−1.751))z_1/z_2 = \sqrt{5}(\cos(-1.751) + i\sin(-1.751)).

UT-2: De Moivre”s Theorem

Section titled “UT-2: De Moivre”s Theorem”

Question: (a) State De Moivre’s theorem. (b) Use De Moivre’s theorem to find (1+i)8(1 + i)^8. (c) Find the four fourth roots of −16-16. (d) Verify that the product of all four roots equals −16-16.

Solution:

(a) De Moivre’s theorem: [r(cos⁡θ+isin⁡θ)]n=rn(cos⁡nθ+isin⁡nθ)[r(\cos\theta + i\sin\theta)]^n = r^n(\cos n\theta + i\sin n\theta).

(b) 1+i=2(cos⁡π/4+isin⁡π/4)1 + i = \sqrt{2}(\cos\pi/4 + i\sin\pi/4). (1+i)8=(2)8(cos⁡2π+isin⁡2π)=16(1+0i)=16(1+i)^8 = (\sqrt{2})^8(\cos 2\pi + i\sin 2\pi) = 16(1 + 0i) = 16.

(c) −16=16(cos⁡π+isin⁡π)-16 = 16(\cos\pi + i\sin\pi). Fourth roots: r = 2$$\theta = \frac{\pi + 2k\pi}{4} for k=0,1,2,3k = 0,1,2,3.

k=0k=0: 2(cos⁡π/4+isin⁡π/4)=2(12+i2)=2+2i2(\cos\pi/4 + i\sin\pi/4) = 2\left(\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = \sqrt{2} + \sqrt{2}i. k=1k=1: 2(cos⁡3π/4+isin⁡3π/4)=2(−12+i2)=−2+2i2(\cos 3\pi/4 + i\sin 3\pi/4) = 2\left(-\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}\right) = -\sqrt{2} + \sqrt{2}i. k=2k=2: 2(cos⁡5π/4+isin⁡5π/4)=2(−12−i2)=−2−2i2(\cos 5\pi/4 + i\sin 5\pi/4) = 2\left(-\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = -\sqrt{2} - \sqrt{2}i. k=3k=3: 2(cos⁡7π/4+isin⁡7π/4)=2(12−i2)=2−2i2(\cos 7\pi/4 + i\sin 7\pi/4) = 2\left(\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}}\right) = \sqrt{2} - \sqrt{2}i.

(d) Product =(2+2i)(−2+2i)(−2−2i)(2−2i)= (\sqrt{2} + \sqrt{2}i)(-\sqrt{2} + \sqrt{2}i)(-\sqrt{2} - \sqrt{2}i)(\sqrt{2} - \sqrt{2}i).

First pair: (2+2i)(−2+2i)=−2+2i−2i+2i2=−2−2=−4(\sqrt{2} + \sqrt{2}i)(-\sqrt{2} + \sqrt{2}i) = -2 + 2i - 2i + 2i^2 = -2 - 2 = -4. Second pair: (−2−2i)(2−2i)=−2+2i−2i+2i2=−2−2=−4(-\sqrt{2} - \sqrt{2}i)(\sqrt{2} - \sqrt{2}i) = -2 + 2i - 2i + 2i^2 = -2 - 2 = -4. Product =(−4)(−4)=16≠−16= (-4)(-4) = 16 \ne -16.

Let me recalculate. Actually: the product of the first pair is (2)2+(2)2=2+2=4(\sqrt{2})^2 + (\sqrt{2})^2 = 2 + 2 = 4… No. (a+bi)(a−bi)=a2+b2(a+bi)(a-bi) = a^2 + b^2. But these are not conjugate pairs in the right order.

Let me just multiply: (−16)1/4(-16)^{1/4} has 4 roots whose product should be the constant term with sign: for z4+16=0z^4 + 16 = 0The product of roots =16= 16 (by Vieta’s, with sign for even degree). This is correct: the product of the 4th roots of −16-16 is (−1)4×16=16(-1)^4 \times 16 = 16… Actually, the product of the nnTh roots of ww is w⋅(−1)n−1w \cdot (-1)^{n-1}. For n=4n=4: (−16)(−1)3=16(-16)(-1)^3 = 16. So the product is 16, not −16-16. The question has an error.

UT-3: Roots of Unity

Section titled “UT-3: Roots of Unity”

Question: (a) Find all five fifth roots of unity and show them on an Argand diagram. (b) Show that the sum of all five roots is zero. (c) Express each root in the form cos⁡(2πk/5)+isin⁡(2πk/5)\cos(2\pi k/5) + i\sin(2\pi k/5) for k=0,1,2,3,4k = 0, 1, 2, 3, 4. (d) Find the value of 1+ω+ω2+ω3+ω41 + \omega + \omega^2 + \omega^3 + \omega^4 where ω=cos⁡(2π/5)+isin⁡(2π/5)\omega = \cos(2\pi/5) + i\sin(2\pi/5).

Solution:

(a) The fifth roots of unity are solutions to z5=1=cos⁡0+isin⁡0z^5 = 1 = \cos 0 + i\sin 0. They lie on the unit circle at angles 2πk/52\pi k/5 for k=0,1,2,3,4k = 0, 1, 2, 3, 4.

(b) The sum of all nnTh roots of unity is zero for n>1n \gt 1. This follows from the fact that they are the roots of zn−1=0z^n - 1 = 0And the coefficient of zn−1z^{n-1} is zero (by Vieta’s formulas).

(c) zk=cos⁡(2πk/5)+isin⁡(2πk/5)z_k = \cos(2\pi k/5) + i\sin(2\pi k/5) for k=0,1,2,3,4k = 0, 1, 2, 3, 4.

z_0 = 1$$z_1 = \cos(2\pi/5) + i\sin(2\pi/5)$$z_2 = \cos(4\pi/5) + i\sin(4\pi/5)$$z_3 = \cos(6\pi/5) + i\sin(6\pi/5)$$z_4 = \cos(8\pi/5) + i\sin(8\pi/5).

(d) Since ω\omega is a primitive 5th root of unity: 1+ω+ω2+ω3+ω4=01 + \omega + \omega^2 + \omega^3 + \omega^4 = 0 (the sum of all 5th roots of unity is zero).

Integration Tests

Section titled “Integration Tests”

IT-1: Complex Numbers and Loci (with Matrices)

Section titled “IT-1: Complex Numbers and Loci (with Matrices)”

Question: (a) Find the locus of points zz such that ∣z−3−i∣=2∣z+1+2i∣|z - 3 - i| = 2|z + 1 + 2i|. Describe the locus geometrically. (b) The transformation w=z−iz+iw = \frac{z - i}{z + i} maps the upper half of the Argand diagram (Im(z)>0\text{Im}(z) \gt 0) to a circle in the ww-plane. Find the centre and radius of this circle. (c) If z=x+iyz = x + iyExpress w=z−iz+iw = \frac{z - i}{z + i} in terms of xx and yy.

Solution:

(a) Let z=x+iyz = x + iy. ∣z−3−i∣2=4∣z+1+2i∣2|z - 3 - i|^2 = 4|z + 1 + 2i|^2. (x−3)2+(y−1)2=4[(x+1)2+(y+2)2](x-3)^2 + (y-1)^2 = 4[(x+1)^2 + (y+2)^2]. x2−6x+9+y2−2y+1=4x2+8x+4+4y2+16y+16x^2 - 6x + 9 + y^2 - 2y + 1 = 4x^2 + 8x + 4 + 4y^2 + 16y + 16. x2−6x+y2−2y+10=4x2+8x+4y2+16y+20x^2 - 6x + y^2 - 2y + 10 = 4x^2 + 8x + 4y^2 + 16y + 20. 0=3x2+14x+3y2+18y+100 = 3x^2 + 14x + 3y^2 + 18y + 10. x2+143x+y2+6y=−103x^2 + \frac{14}{3}x + y^2 + 6y = -\frac{10}{3}. (x+73)2−499+(y+3)2−9=−103\left(x + \frac{7}{3}\right)^2 - \frac{49}{9} + (y+3)^2 - 9 = -\frac{10}{3}. (x+73)2+(y+3)2=499+9−103=49+81−309=1009\left(x + \frac{7}{3}\right)^2 + (y+3)^2 = \frac{49}{9} + 9 - \frac{10}{3} = \frac{49 + 81 - 30}{9} = \frac{100}{9}.

The locus is a circle with centre (−73,−3)\left(-\frac{7}{3}, -3\right) and radius 103\frac{10}{3}.

(c) w=x+iy−ix+iy+i=x+i(y−1)x+i(y+1)=[x+i(y−1)][x−i(y+1)]x2+(y+1)2w = \frac{x + iy - i}{x + iy + i} = \frac{x + i(y-1)}{x + i(y+1)} = \frac{[x + i(y-1)][x - i(y+1)]}{x^2 + (y+1)^2}.

Numerator: x2−ix(y+1)+ix(y−1)+(y−1)(y+1)=x2+(y2−1)+i[−x(y+1)+x(y−1)]=x2+y2−1−2ixx^2 - ix(y+1) + ix(y-1) + (y-1)(y+1) = x^2 + (y^2 - 1) + i[-x(y+1) + x(y-1)] = x^2 + y^2 - 1 - 2ix.

w=x2+y2−1−2ixx2+(y+1)2w = \frac{x^2 + y^2 - 1 - 2ix}{x^2 + (y+1)^2}.

IT-2: De Moivre and Trigonometry (with Further Algebra)

Section titled “IT-2: De Moivre and Trigonometry (with Further Algebra)”

Question: (a) Use De Moivre’s theorem to find expressions for cos⁡3θ\cos 3\theta and sin⁡3θ\sin 3\theta in terms of cos⁡θ\cos\theta and sin⁡θ\sin\theta. (b) Hence solve cos⁡3θ=0\cos 3\theta = 0 for 0≤θ≤π0 \le \theta \le \pi. (c) Find the exact value of cos⁡3θ\cos 3\theta when cos⁡θ=1/3\cos\theta = 1/3. (d) Show that cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = 4\cos^3\theta - 3\cos\theta.

Solution:

(a) (cos⁡θ+isin⁡θ)3=cos⁡3θ+isin⁡3θ(\cos\theta + i\sin\theta)^3 = \cos 3\theta + i\sin 3\theta.

Expanding using the binomial theorem: =cos⁡3θ+3cos⁡2θ(isin⁡θ)+3cos⁡θ(isin⁡θ)2+(isin⁡θ)3= \cos^3\theta + 3\cos^2\theta(i\sin\theta) + 3\cos\theta(i\sin\theta)^2 + (i\sin\theta)^3 =cos⁡3θ+3icos⁡2θsin⁡θ−3cos⁡θsin⁡2θ−isin⁡3θ= \cos^3\theta + 3i\cos^2\theta\sin\theta - 3\cos\theta\sin^2\theta - i\sin^3\theta =(cos⁡3θ−3cos⁡θsin⁡2θ)+i(3cos⁡2θsin⁡θ−sin⁡3θ)= (\cos^3\theta - 3\cos\theta\sin^2\theta) + i(3\cos^2\theta\sin\theta - \sin^3\theta).

cos⁡3θ=cos⁡3θ−3cos⁡θsin⁡2θ\cos 3\theta = \cos^3\theta - 3\cos\theta\sin^2\theta. sin⁡3θ=3cos⁡2θsin⁡θ−sin⁡3θ\sin 3\theta = 3\cos^2\theta\sin\theta - \sin^3\theta.

(b) cos⁡3θ=0\cos 3\theta = 0 means 3θ=π/2,3π/2,5π/2,...3\theta = \pi/2, 3\pi/2, 5\pi/2, ... So θ=π/6,π/2,5π/6\theta = \pi/6, \pi/2, 5\pi/6 in [0,π][0, \pi].

(c) cos⁡3θ=4cos⁡3θ−3cos⁡θ=4(1/27)−3(1/3)=4/27−9/9=4/27−1=−23/27\cos 3\theta = 4\cos^3\theta - 3\cos\theta = 4(1/27) - 3(1/3) = 4/27 - 9/9 = 4/27 - 1 = -23/27.

(d) Using sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta: cos⁡3θ=cos⁡3θ−3cos⁡θ(1−cos⁡2θ)=cos⁡3θ−3cos⁡θ+3cos⁡3θ=4cos⁡3θ−3cos⁡θ\cos 3\theta = \cos^3\theta - 3\cos\theta(1 - \cos^2\theta) = \cos^3\theta - 3\cos\theta + 3\cos^3\theta = 4\cos^3\theta - 3\cos\theta.

IT-3: Complex Transformations (with Polar Coordinates)

Section titled “IT-3: Complex Transformations (with Polar Coordinates)”

Question: (a) The transformation w=z2w = z^2 maps the line Re(z)=1\text{Re}(z) = 1 in the zz-plane. Find the image in the ww-plane. (b) If z=2eiπ/6z = 2e^{i\pi/6}Find w=z4w = z^4 in modulus-argument form. (c) Find the image of the region |z| \le 2$$0 \le \arg(z) \le \pi/4 under the transformation w=z3w = z^3. (d) A complex number zz satisfies ∣z−1−2i∣=∣z−3−4i∣|z - 1 - 2i| = |z - 3 - 4i|. Show that this represents a straight line and find its equation.

Solution:

(a) Let z=1+iyz = 1 + iy. w=z2=(1+iy)2=1−y2+2iyw = z^2 = (1+iy)^2 = 1 - y^2 + 2iy. Let w=u+ivw = u + iv: u=1−y2u = 1 - y^2 and v=2yv = 2y. From v=2yv = 2y: y=v/2y = v/2. u=1−v2/4u = 1 - v^2/4. This is a parabola u=1−v2/4u = 1 - v^2/4Opening to the left.

(b) w=z4=(2eiπ/6)4=24ei4π/6=16ei2π/3=16(cos⁡2π/3+isin⁡2π/3)=16(−1/2+i3/2)=−8+83iw = z^4 = (2e^{i\pi/6})^4 = 2^4 e^{i4\pi/6} = 16e^{i2\pi/3} = 16(\cos 2\pi/3 + i\sin 2\pi/3) = 16(-1/2 + i\sqrt{3}/2) = -8 + 8\sqrt{3}i.

(c) Under w=z3w = z^3: ∣w∣=∣z∣3≤8|w| = |z|^3 \le 8And arg⁡(w)=3arg⁡(z)∈[0,3π/4]\arg(w) = 3\arg(z) \in [0, 3\pi/4]. The image is the sector |w| \le 8$$0 \le \arg(w) \le 3\pi/4.

(d) ∣z−1−2i∣2=∣z−3−4i∣2|z - 1 - 2i|^2 = |z - 3 - 4i|^2. Let z=x+iyz = x + iy: (x−1)2+(y−2)2=(x−3)2+(y−4)2(x-1)^2 + (y-2)^2 = (x-3)^2 + (y-4)^2. x2−2x+1+y2−4y+4=x2−6x+9+y2−8y+16x^2 - 2x + 1 + y^2 - 4y + 4 = x^2 - 6x + 9 + y^2 - 8y + 16. −2x−4y+5=−6x−8y+25-2x - 4y + 5 = -6x - 8y + 25. 4x+4y=204x + 4y = 20. x+y=5x + y = 5.

This is a straight line: the perpendicular bisector of the segment joining (1,2)(1, 2) and (3,4)(3, 4).

Common Mistakes

Section titled “Common Mistakes”

Confusing the modulus and argument when converting between forms: When converting z=a+biz = a + bi to modulus-argument form, the modulus is ∣z∣=a2+b2|z| = \sqrt{a^2 + b^2} and the argument is arg⁡(z)=arctan⁡(b/a)\arg(z) = \arctan(b/a) (adjusted for the correct quadrant). Students often forget to adjust the argument for quadrants 2 and 3, giving the wrong angle.

Forgetting that i2=−1i^2 = -1 when expanding powers: When calculating (a+bi)2(a + bi)^2 or higher powers, students frequently make sign errors by forgetting that i2=−1i^2 = -1, i3=−ii^3 = -i, and i4=1i^4 = 1. Write out each power of ii explicitly until the pattern is automatic.

Confusing the complex conjugate with the negative: The conjugate of z=a+biz = a + bi is zˉ=a−bi\bar{z} = a - bi (flip the sign of the imaginary part). The negative is −z=−a−bi-z = -a - bi (flip both parts). Students sometimes compute −z-z when asked for zˉ\bar{z}, or vice versa. These are different operations.

Cross-References

Section titled “Cross-References”