IB Mathematics: Analysis and Approaches Quiz: Polar And Exponential Form
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Polar And Exponential FormQuestion 1 of 20

If z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta, which of the following expressions is equivalent to zn+z−nz^n + z^{-n}?

cos⁡(nθ)\cos(n\theta)
2cos⁡(nθ)2\cos(n\theta)
2isin⁡(nθ)2i\sin(n\theta)
2cos⁡nθ2\cos^n\theta
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Polar And Exponential Form

Practice Polar And Exponential Form in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Polar And Exponential Form, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Question 1

If z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta, which of the following expressions is equivalent to zn+z−nz^n + z^{-n}?

  1. cos⁡(nθ)\cos(n\theta)
  2. 2cos⁡(nθ)2\cos(n\theta) (correct answer)
  3. 2isin⁡(nθ)2i\sin(n\theta)
  4. 2cos⁡nθ2\cos^n\theta
Explanation: Given z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta, we can use De Moivre's theorem to find znz^n and z−nz^{-n}. According to the theorem, zn=(cos⁡θ+isin⁡θ)n=cos⁡(nθ)+isin⁡(nθ)z^n = (\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta). Similarly, z−n=(cos⁡θ+isin⁡θ)−n=cos⁡(−nθ)+isin⁡(−nθ)z^{-n} = (\cos\theta + i\sin\theta)^{-n} = \cos(-n\theta) + i\sin(-n\theta). Using the properties of trigonometric functions, cos⁡(−x)=cos⁡(x)\cos(-x) = \cos(x) and sin⁡(−x)=−sin⁡(x)\sin(-x) = -\sin(x), we can simplify z−nz^{-n} to cos⁡(nθ)−isin⁡(nθ)\cos(n\theta) - i\sin(n\theta). Now, we add the two expressions: zn+z−n=(cos⁡(nθ)+isin⁡(nθ))+(cos⁡(nθ)−isin⁡(nθ))z^n + z^{-n} = (\cos(n\theta) + i\sin(n\theta)) + (\cos(n\theta) - i\sin(n\theta)). The imaginary parts cancel out, leaving 2cos⁡(nθ)2\cos(n\theta).

Question 2

Given that z1=r1eiθ1z_1 = r_1 e^{i\theta_1} and z2=r2eiθ2z_2 = r_2 e^{i\theta_2} are such that z1z2z_1 z_2 is purely real and z1z2\frac{z_1}{z_2} is purely imaginary, which of the following is a possible value for θ1\theta_1?

  1. −π4-\frac{\pi}{4} (correct answer)
  2. 0
  3. π3\frac{\pi}{3}
  4. π2\frac{\pi}{2}
Explanation: The product is z1z2=r1r2ei(θ1+θ2)z_1 z_2 = r_1 r_2 e^{i(\theta_1+\theta_2)}. For this to be purely real, its argument must be a multiple of π\pi. So, θ1+θ2=kπ\theta_1+\theta_2 = k\pi for some integer kk. The quotient is z1z2=r1r2ei(θ1−θ2)\frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i(\theta_1-\theta_2)}. For this to be purely imaginary, its argument must be an odd multiple of π2\frac{\pi}{2}. So, θ1−θ2=(2m+1)π2\theta_1-\theta_2 = \frac{(2m+1)\pi}{2} for some integer mm. We have a system of two equations: (1) θ1+θ2=kπ\theta_1+\theta_2 = k\pi and (2) θ1−θ2=(m+12)π\theta_1-\theta_2 = (m+\frac{1}{2})\pi. Adding the two equations gives 2θ1=kπ+(m+12)π=(k+m+12)π2\theta_1 = k\pi + (m+\frac{1}{2})\pi = (k+m+\frac{1}{2})\pi. Therefore, θ1=(k+m+12)π2=(2(k+m)+1)π4\theta_1 = \frac{(k+m+\frac{1}{2})\pi}{2} = \frac{(2(k+m)+1)\pi}{4}. This means θ1\theta_1 must be an odd multiple of π4\frac{\pi}{4}. Among the given options, only −π4-\frac{\pi}{4} fits this condition (for 2(k+m)+1=−12(k+m)+1 = -1).

Question 3

Given two non-zero complex numbers z1z_1 and z2z_2 with principal arguments arg⁡(z1)=α\arg(z_1) = \alpha and arg⁡(z2)=β\arg(z_2) = \beta, where π2<α<π\frac{\pi}{2} < \alpha < \pi and −π2<β<0-\frac{\pi}{2} < \beta < 0. Which of the following statements must be true about their product z1z2z_1 z_2?

  1. Re⁡(z1z2)>0\operatorname{Re}(z_1 z_2) > 0
  2. Re⁡(z1z2)<0\operatorname{Re}(z_1 z_2) < 0
  3. Im⁡(z1z2)>0\operatorname{Im}(z_1 z_2) > 0 (correct answer)
  4. Im⁡(z1z2)<0\operatorname{Im}(z_1 z_2) < 0
Explanation: The argument of the product is the sum of the arguments: arg⁡(z1z2)=arg⁡(z1)+arg⁡(z2)=α+β\arg(z_1 z_2) = \arg(z_1) + \arg(z_2) = \alpha + \beta. We need to find the range of α+β\alpha + \beta. Given π2<α<π\frac{\pi}{2} < \alpha < \pi and −π2<β<0-\frac{\pi}{2} < \beta < 0. The minimum value of the sum is π2+(−π2)=0\frac{\pi}{2} + (-\frac{\pi}{2}) = 0. The maximum value of the sum is π+0=π\pi + 0 = \pi. Therefore, 0<α+β<π0 < \alpha + \beta < \pi. Let θ=arg⁡(z1z2)\theta = \arg(z_1 z_2). We have 0<θ<π0 < \theta < \pi. For any complex number w=∣w∣(%cosθ+isin⁡θ)w = |w|(\%cos\theta + i\sin\theta), its imaginary part is ∣w∣sin⁡θ|w|\sin\theta. Since z1z_1 and z2z_2 are non-zero, ∣w∣>0|w| > 0. In the interval 0<θ<π0 < \theta < \pi, sin⁡θ>0\sin\theta > 0. Thus, Im⁡(z1z2)>0\operatorname{Im}(z_1 z_2) > 0. The real part, ∣w∣cos⁡θ|w|\cos\theta, can be positive (if 0<θ<π/20 < \theta < \pi/2) or negative (if π/2<θ<π\pi/2 < \theta < \pi), so we cannot determine the sign of the real part.

Question 4

Let ω\omega be a non-real cube root of unity. What is the value of the expression (1−ω+ω2)(1+ω−ω2)(1 - \omega + \omega^2)(1 + \omega - \omega^2)?

  1. -4
  2. 0
  3. 3
  4. 4 (correct answer)
Explanation: The cube roots of unity are the solutions to z3−1=0z^3 - 1 = 0, which factors as (z−1)(z2+z+1)=0(z-1)(z^2+z+1) = 0. Since ω\omega is a non-real root, it must be a root of z2+z+1=0z^2+z+1=0. This gives us the crucial identity 1+ω+ω2=01+\omega+\omega^2=0. From this, we can derive 1+ω2=−ω1+\omega^2 = -\omega and 1+ω=−ω21+\omega = -\omega^2. Now we simplify the given expression. The first factor is 1−ω+ω2=(1+ω2)−ω=−ω−ω=−2ω1 - \omega + \omega^2 = (1+\omega^2) - \omega = -\omega - \omega = -2\omega. The second factor is 1+ω−ω2=(1+ω)−ω2=−ω2−ω2=−2ω21 + \omega - \omega^2 = (1+\omega) - \omega^2 = -\omega^2 - \omega^2 = -2\omega^2. The product is (−2ω)(−2ω2)=4ω3(-2\omega)(-2\omega^2) = 4\omega^3. Since ω\omega is a cube root of unity, ω3=1\omega^3 = 1. Therefore, the value of the expression is 4(1)=44(1) = 4.

Question 5

The three distinct cube roots of z=8iz = 8i form the vertices of a triangle in the Argand plane. What is the area of this triangle?

  1. 3
  2. 333\sqrt{3} (correct answer)
  3. 6
  4. 434\sqrt{3}
Explanation: First, express z=8iz=8i in polar form. The modulus is r=∣8i∣=8r=|8i|=8. The argument is θ=π2\theta = \frac{\pi}{2}. So z=8eiπ/2z = 8e^{i\pi/2}. The cube roots wkw_k are given by wk=83ei(π/2+2kπ3)w_k = \sqrt[3]{8} e^{i(\frac{\pi/2 + 2k\pi}{3})} for k=0,1,2k=0, 1, 2. The modulus of each root is R=83=2R = \sqrt[3]{8} = 2. The roots lie on a circle of radius 2 centered at the origin. The three cube roots form the vertices of an equilateral triangle inscribed in this circle. The area of a regular n-gon inscribed in a circle of radius R is given by A=12nR2sin⁡(2πn)A = \frac{1}{2}nR^2\sin(\frac{2\pi}{n}). For an equilateral triangle, n=3n=3, so the area is A=12(3)(22)sin⁡(2π3)=32(4)(32)=33A = \frac{1}{2}(3)(2^2)\sin(\frac{2\pi}{3}) = \frac{3}{2}(4)(\frac{\sqrt{3}}{2}) = 3\sqrt{3}.

Question 6

Given that z1=r1eiθ1z_1 = r_1 e^{i\theta_1} and z2=r2eiθ2z_2 = r_2 e^{i\theta_2} are such that z1z2z_1 z_2 is purely real and z1z2\frac{z_1}{z_2} is purely imaginary, which of the following is a possible value for θ1\theta_1?

  1. −π4-\frac{\pi}{4} (correct answer)
  2. 0
  3. π3\frac{\pi}{3}
  4. π2\frac{\pi}{2}
Explanation: The product is z1z2=r1r2ei(θ1+θ2)z_1 z_2 = r_1 r_2 e^{i(\theta_1+\theta_2)}. For this to be purely real, its argument must be a multiple of π\pi. So, θ1+θ2=kπ\theta_1+\theta_2 = k\pi for some integer kk. The quotient is z1z2=r1r2ei(θ1−θ2)\frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i(\theta_1-\theta_2)}. For this to be purely imaginary, its argument must be an odd multiple of π2\frac{\pi}{2}. So, θ1−θ2=(2m+1)π2\theta_1-\theta_2 = \frac{(2m+1)\pi}{2} for some integer mm. We have a system of two equations: (1) θ1+θ2=kπ\theta_1+\theta_2 = k\pi and (2) θ1−θ2=(m+12)π\theta_1-\theta_2 = (m+\frac{1}{2})\pi. Adding the two equations gives 2θ1=kπ+(m+12)π=(k+m+12)π2\theta_1 = k\pi + (m+\frac{1}{2})\pi = (k+m+\frac{1}{2})\pi. Therefore, θ1=(k+m+12)π2=(2(k+m)+1)π4\theta_1 = \frac{(k+m+\frac{1}{2})\pi}{2} = \frac{(2(k+m)+1)\pi}{4}. This means θ1\theta_1 must be an odd multiple of π4\frac{\pi}{4}. Among the given options, only −π4-\frac{\pi}{4} fits this condition (for 2(k+m)+1=−12(k+m)+1 = -1).

Question 7

Let the complex number zz be represented by z=(3+i)5(1−i)8z = \frac{(\sqrt{3}+i)^5}{(1-i)^8}. Find the principal argument of zz.

  1. −5π6-\frac{5\pi}{6}
  2. −π6-\frac{\pi}{6}
  3. π6\frac{\pi}{6}
  4. 5π6\frac{5\pi}{6} (correct answer)
Explanation: First, convert the numerator and denominator into polar (exponential) form. For the numerator, let z1=3+iz_1 = \sqrt{3}+i. The modulus is ∣z1∣=(3)2+12=2|z_1| = \sqrt{(\sqrt{3})^2+1^2} = 2. The argument is arg⁡(z1)=arctan⁡(1/3)=π6\arg(z_1) = \arctan(1/\sqrt{3}) = \frac{\pi}{6}. So, z1=2eiπ/6z_1 = 2e^{i\pi/6}. For the denominator, let z2=1−iz_2 = 1-i. The modulus is ∣z2∣=12+(−1)2=2|z_2| = \sqrt{1^2+(-1)^2} = \sqrt{2}. The argument is arg⁡(z2)=arctan⁡(−1/1)=−π4\arg(z_2) = \arctan(-1/1) = -\frac{\pi}{4}. So, z2=2e−iπ/4z_2 = \sqrt{2}e^{-i\pi/4}. Now, use De Moivre's theorem. z15=(2eiπ/6)5=32ei5π/6z_1^5 = (2e^{i\pi/6})^5 = 32e^{i5\pi/6}. z28=(2e−iπ/4)8=(2)8e−i(8π/4)=16e−i2π=16ei0z_2^8 = (\sqrt{2}e^{-i\pi/4})^8 = (\sqrt{2})^8 e^{-i(8\pi/4)} = 16e^{-i2\pi} = 16e^{i0}. Then, z=32ei5π/616ei0=2ei5π/6z = \frac{32e^{i5\pi/6}}{16e^{i0}} = 2e^{i5\pi/6}. The principal argument of zz is 5π6\frac{5\pi}{6}.

Question 8

Given two non-zero complex numbers z1z_1 and z2z_2 with principal arguments arg⁡(z1)=α\arg(z_1) = \alpha and arg⁡(z2)=β\arg(z_2) = \beta, where π2<α<π\frac{\pi}{2} < \alpha < \pi and −π2<β<0-\frac{\pi}{2} < \beta < 0. Which of the following statements must be true about their product z1z2z_1 z_2?

  1. Re⁡(z1z2)>0\operatorname{Re}(z_1 z_2) > 0
  2. Re⁡(z1z2)<0\operatorname{Re}(z_1 z_2) < 0
  3. Im⁡(z1z2)>0\operatorname{Im}(z_1 z_2) > 0 (correct answer)
  4. Im⁡(z1z2)<0\operatorname{Im}(z_1 z_2) < 0
Explanation: The argument of the product is the sum of the arguments: arg⁡(z1z2)=arg⁡(z1)+arg⁡(z2)=α+β\arg(z_1 z_2) = \arg(z_1) + \arg(z_2) = \alpha + \beta. We need to find the range of α+β\alpha + \beta. Given π2<α<π\frac{\pi}{2} < \alpha < \pi and −π2<β<0-\frac{\pi}{2} < \beta < 0. The minimum value of the sum is π2+(−π2)=0\frac{\pi}{2} + (-\frac{\pi}{2}) = 0. The maximum value of the sum is π+0=π\pi + 0 = \pi. Therefore, 0<α+β<π0 < \alpha + \beta < \pi. Let θ=arg⁡(z1z2)\theta = \arg(z_1 z_2). We have 0<θ<π0 < \theta < \pi. For any complex number w=∣w∣(%cosθ+isin⁡θ)w = |w|(\%cos\theta + i\sin\theta), its imaginary part is ∣w∣sin⁡θ|w|\sin\theta. Since z1z_1 and z2z_2 are non-zero, ∣w∣>0|w| > 0. In the interval 0<θ<π0 < \theta < \pi, sin⁡θ>0\sin\theta > 0. Thus, Im⁡(z1z2)>0\operatorname{Im}(z_1 z_2) > 0. The real part, ∣w∣cos⁡θ|w|\cos\theta, can be positive (if 0<θ<π/20 < \theta < \pi/2) or negative (if π/2<θ<π\pi/2 < \theta < \pi), so we cannot determine the sign of the real part.

Question 9

Let z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta. For what value of θ\theta in the range 0≤θ<2π0 \le \theta < 2\pi is the real part of 11−z\frac{1}{1-z} equal to 12\frac{1}{2}?

  1. θ=π3,5π3\theta = \frac{\pi}{3}, \frac{5\pi}{3}
  2. θ=π2,3π2\theta = \frac{\pi}{2}, \frac{3\pi}{2}
  3. θ=π4,7π4\theta = \frac{\pi}{4}, \frac{7\pi}{4}
  4. Any θ≠0\theta \neq 0 (correct answer)
Explanation: Let z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta. We analyze the expression 11−z\frac{1}{1-z}. Substitute for z: 11−(cos⁡θ+isin⁡θ)=1(1−cos⁡θ)−isin⁡θ\frac{1}{1 - (\cos\theta + i\sin\theta)} = \frac{1}{(1-\cos\theta) - i\sin\theta}. To find the real part, multiply the numerator and denominator by the conjugate of the denominator: (1−cos⁡θ)+isin⁡θ((1−cos⁡θ)−isin⁡θ)((1−cos⁡θ)+isin⁡θ)=1−cos⁡θ+isin⁡θ(1−cos⁡θ)2+sin⁡2θ\frac{(1-\cos\theta) + i\sin\theta}{((1-\cos\theta) - i\sin\theta)((1-\cos\theta) + i\sin\theta)} = \frac{1-\cos\theta + i\sin\theta}{(1-\cos\theta)^2 + \sin^2\theta}. Let's simplify the denominator: 1−2cos⁡θ+cos⁡2θ+sin⁡2θ=1−2cos⁡θ+1=2−2cos⁡θ=2(1−cos⁡θ)1 - 2\cos\theta + \cos^2\theta + \sin^2\theta = 1 - 2\cos\theta + 1 = 2 - 2\cos\theta = 2(1-\cos\theta). The expression becomes 1−cos⁡θ+isin⁡θ2(1−cos⁡θ)\frac{1-\cos\theta + i\sin\theta}{2(1-\cos\theta)}. The real part is Re⁡(11−z)=1−cos⁡θ2(1−cos⁡θ)\operatorname{Re}\left(\frac{1}{1-z}\right) = \frac{1-\cos\theta}{2(1-\cos\theta)}. As long as 1−cos⁡θ≠01-\cos\theta \neq 0 (i.e., cos⁡θ≠1\cos\theta \neq 1, which means θ≠0\theta \neq 0 in the given range), we can cancel the term 1−cos⁡θ1-\cos\theta, and the real part is 12\frac{1}{2}. Therefore, the condition is met for any θ\theta in the range 0≤θ<2π0 \le \theta < 2\pi except for θ=0\theta=0.

Question 10

Let z=22+i22z = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}. If a complex number ww is multiplied by z3z^3, what is the geometric transformation applied to ww in the Argand diagram?

  1. Rotation by 45∘45^\circ counter-clockwise about the origin
  2. Rotation by 135∘135^\circ clockwise about the origin
  3. Rotation by 135∘135^\circ counter-clockwise about the origin (correct answer)
  4. Scaling by 322\frac{3\sqrt{2}}{2} and rotation by 135∘135^\circ counter-clockwise
Explanation: First, we express zz in polar form. The modulus is ∣z∣=(22)2+(22)2=24+24=1=1|z| = \sqrt{(\frac{\sqrt{2}}{2})^2 + (\frac{\sqrt{2}}{2})^2} = \sqrt{\frac{2}{4} + \frac{2}{4}} = \sqrt{1} = 1. The argument is arg⁡(z)=arctan⁡(2/22/2)=arctan⁡(1)=π4\arg(z) = \arctan\left(\frac{\sqrt{2}/2}{\sqrt{2}/2}\right) = \arctan(1) = \frac{\pi}{4} or 45∘45^\circ. So, z=eiπ/4z = e^{i\pi/4}. Next, we find z3z^3 using De Moivre's theorem: z3=(eiπ/4)3=ei3π/4z^3 = (e^{i\pi/4})^3 = e^{i3\pi/4}. The modulus of z3z^3 is 1 and its argument is 3π4\frac{3\pi}{4}, which is 135∘135^\circ. Multiplying a complex number ww by a complex number ρeiθ\rho e^{i\theta} corresponds to scaling ww by a factor of ρ\rho and rotating it counter-clockwise by an angle of θ\theta about the origin. Here, z3z^3 has modulus ρ=1\rho=1 and argument θ=135∘\theta = 135^\circ. Thus, the transformation is a rotation by 135∘135^\circ counter-clockwise about the origin with no scaling.

Question 11

Let z=eiθz = e^{i\theta}. The expression z2−1z2+1\frac{z^2 - 1}{z^2 + 1} simplifies to which of the following?

  1. tan⁡θ\tan\theta
  2. icot⁡θi\cot\theta
  3. itan⁡θi\tan\theta (correct answer)
  4. itan⁡(2θ)i\tan(2\theta)
Explanation: Let z=eiθz=e^{i\theta}. Then z2=ei2θz^2 = e^{i2\theta}. Substitute this into the expression: ei2θ−1ei2θ+1\frac{e^{i2\theta} - 1}{e^{i2\theta} + 1}. To simplify, multiply the numerator and denominator by e−iθe^{-i\theta}: e−iθ(ei2θ−1)e−iθ(ei2θ+1)=eiθ−e−iθeiθ+e−iθ\frac{e^{-i\theta}(e^{i2\theta} - 1)}{e^{-i\theta}(e^{i2\theta} + 1)} = \frac{e^{i\theta} - e^{-i\theta}}{e^{i\theta} + e^{-i\theta}}. Using Euler's relations, cos⁡θ=eiθ+e−iθ2\cos\theta = \frac{e^{i\theta}+e^{-i\theta}}{2} and sin⁡θ=eiθ−e−iθ2i\sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}, we can rewrite the numerator as 2isin⁡θ2i\sin\theta and the denominator as 2cos⁡θ2\cos\theta. The expression becomes 2isin⁡θ2cos⁡θ=itan⁡θ\frac{2i\sin\theta}{2\cos\theta} = i\tan\theta.

Question 12

A complex number zz has modulus rr and argument θ\theta. The complex number w=zzˉ−zˉzw = \frac{z}{\bar{z}} - \frac{\bar{z}}{z} is purely imaginary. What is its imaginary part?

  1. 2sin⁡(2θ)2\sin(2\theta) (correct answer)
  2. 2isin⁡(2θ)2i\sin(2\theta)
  3. 2cos⁡(2θ)2\cos(2\theta)
  4. sin⁡(2θ)\sin(2\theta)
Explanation: Let z=reiθz = re^{i\theta}. Then its conjugate is zˉ=re−iθ\bar{z} = re^{-i\theta}. Now we can evaluate the terms in the expression for ww. zzˉ=reiθre−iθ=eiθ−(−iθ)=ei2θ\frac{z}{\bar{z}} = \frac{re^{i\theta}}{re^{-i\theta}} = e^{i\theta - (-i\theta)} = e^{i2\theta}. zˉz=re−iθreiθ=e−iθ−iθ=e−i2θ\frac{\bar{z}}{z} = \frac{re^{-i\theta}}{re^{i\theta}} = e^{-i\theta - i\theta} = e^{-i2\theta}. So, w=ei2θ−e−i2θw = e^{i2\theta} - e^{-i2\theta}. Using Euler's formula, eiϕ=cos⁡ϕ+isin⁡ϕe^{i\phi} = \cos\phi + i\sin\phi, we have w=(cos⁡(2θ)+isin⁡(2θ))−(cos⁡(−2θ)+isin⁡(−2θ))w = (\cos(2\theta) + i\sin(2\theta)) - (\cos(-2\theta) + i\sin(-2\theta)). This simplifies to w=(cos⁡(2θ)+isin⁡(2θ))−(cos⁡(2θ)−isin⁡(2θ))=2isin⁡(2θ)w = (\cos(2\theta) + i\sin(2\theta)) - (\cos(2\theta) - i\sin(2\theta)) = 2i\sin(2\theta). This is a purely imaginary number. The imaginary part is the coefficient of ii, which is 2sin⁡(2θ)2\sin(2\theta). Note that 2isin⁡(2θ)2i\sin(2\theta) is the complex number itself, not its imaginary part.

Question 13

The locus of a point zz in the complex plane is given by ∣z−4∣=∣z−2i∣|z - 4| = |z - 2i|. Find the argument of the point on this locus that is closest to the origin.

  1. arctan⁡(−2)\arctan(-2)
  2. arctan⁡(−1/2)\arctan(-1/2) (correct answer)
  3. arctan⁡(1/2)\arctan(1/2)
  4. arctan⁡(2)\arctan(2)
Explanation: The equation ∣z−z1∣=∣z−z2∣|z-z_1| = |z-z_2| describes the perpendicular bisector of the line segment connecting z1z_1 and z2z_2. Here, z1=4z_1 = 4 and z2=2iz_2 = 2i. The segment connects the points (4, 0) and (0, 2). The midpoint of this segment is (4+02,0+22)=(2,1)\left(\frac{4+0}{2}, \frac{0+2}{2}\right) = (2, 1). The slope of the segment is 2−00−4=−12\frac{2-0}{0-4} = -\frac{1}{2}. The slope of the perpendicular bisector is the negative reciprocal, which is 2. The equation of the locus is the line passing through (2, 1) with slope 2: y−1=2(x−2)⇒y=2x−3y-1 = 2(x-2) \Rightarrow y = 2x - 3. The point on this line closest to the origin is the foot of the perpendicular from the origin to the line. The line from the origin perpendicular to the locus has a slope of −12-\frac{1}{2}, so its equation is y=−12xy = -\frac{1}{2}x. To find the intersection point, we set the y-values equal: 2x−3=−12x⇒4x−6=−x⇒5x=6⇒x=652x - 3 = -\frac{1}{2}x \Rightarrow 4x - 6 = -x \Rightarrow 5x = 6 \Rightarrow x = \frac{6}{5}. Then y=−12(65)=−35y = -\frac{1}{2}(\frac{6}{5}) = -\frac{3}{5}. The point is z=65−35iz = \frac{6}{5} - \frac{3}{5}i. The argument is arg⁡(z)=arctan⁡(−3/56/5)=arctan⁡(−12)\arg(z) = \arctan\left(\frac{-3/5}{6/5}\right) = \arctan(-\frac{1}{2}).

Question 14

What is the argument of the complex number z=1−cos⁡(2α)−isin⁡(2α)z = 1 - \cos(2\alpha) - i\sin(2\alpha) for 0<α<π/20 < \alpha < \pi/2?

  1. 2α2\alpha
  2. α\alpha
  3. α−π2\alpha - \frac{\pi}{2} (correct answer)
  4. π2−α\frac{\pi}{2} - \alpha
Explanation: We can factor the expression for zz. Using half-angle identities, 1−cos⁡(2α)=2sin⁡2α1-\cos(2\alpha) = 2\sin^2\alpha and sin⁡(2α)=2sin⁡αcos⁡α\sin(2\alpha) = 2\sin\alpha\cos\alpha. Substitute these into zz: z=2sin⁡2α−i(2sin⁡αcos⁡α)z = 2\sin^2\alpha - i(2\sin\alpha\cos\alpha). Since 0<α<π/20 < \alpha < \pi/2, sin⁡α>0\sin\alpha > 0, so we can factor out 2sin⁡α2\sin\alpha: z=2sin⁡α(sin⁡α−icos⁡α)z = 2\sin\alpha(\sin\alpha - i\cos\alpha). The modulus is r=2sin⁡αr = 2\sin\alpha. The argument is determined by the term in the parenthesis. Let's convert this to polar form. We want to write sin⁡α−icos⁡α\sin\alpha - i\cos\alpha as cos⁡θ+isin⁡θ\cos\theta + i\sin\theta. We can use the identities sin⁡α=cos⁡(π2−α)\sin\alpha = \cos(\frac{\pi}{2}-\alpha) and cos⁡α=sin⁡(π2−α)\cos\alpha = \sin(\frac{\pi}{2}-\alpha). So, sin⁡α−icos⁡α=cos⁡(π2−α)−isin⁡(π2−α)\sin\alpha - i\cos\alpha = \cos(\frac{\pi}{2}-\alpha) - i\sin(\frac{\pi}{2}-\alpha). This is equal to cos⁡(−(π2−α))+isin⁡(−(π2−α))=cos⁡(α−π2)+isin⁡(α−π2)\cos(-(\frac{\pi}{2}-\alpha)) + i\sin(-(\frac{\pi}{2}-\alpha)) = \cos(\alpha-\frac{\pi}{2}) + i\sin(\alpha-\frac{\pi}{2}). Thus, the argument of zz is α−π2\alpha - \frac{\pi}{2}.

Question 15

The three distinct cube roots of z=8iz = 8i form the vertices of a triangle in the Argand plane. What is the area of this triangle?

  1. 3
  2. 333\sqrt{3} (correct answer)
  3. 6
  4. 434\sqrt{3}
Explanation: First, express z=8iz=8i in polar form. The modulus is r=∣8i∣=8r=|8i|=8. The argument is θ=π2\theta = \frac{\pi}{2}. So z=8eiπ/2z = 8e^{i\pi/2}. The cube roots wkw_k are given by wk=83ei(π/2+2kπ3)w_k = \sqrt[3]{8} e^{i(\frac{\pi/2 + 2k\pi}{3})} for k=0,1,2k=0, 1, 2. The modulus of each root is R=83=2R = \sqrt[3]{8} = 2. The roots lie on a circle of radius 2 centered at the origin. The three cube roots form the vertices of an equilateral triangle inscribed in this circle. The area of a regular n-gon inscribed in a circle of radius R is given by A=12nR2sin⁡(2πn)A = \frac{1}{2}nR^2\sin(\frac{2\pi}{n}). For an equilateral triangle, n=3n=3, so the area is A=12(3)(22)sin⁡(2π3)=32(4)(32)=33A = \frac{1}{2}(3)(2^2)\sin(\frac{2\pi}{3}) = \frac{3}{2}(4)(\frac{\sqrt{3}}{2}) = 3\sqrt{3}.

Question 16

Let z=eiθz = e^{i\theta}. The expression z2−1z2+1\frac{z^2 - 1}{z^2 + 1} simplifies to which of the following?

  1. tan⁡θ\tan\theta
  2. icot⁡θi\cot\theta
  3. itan⁡θi\tan\theta (correct answer)
  4. itan⁡(2θ)i\tan(2\theta)
Explanation: Let z=eiθz=e^{i\theta}. Then z2=ei2θz^2 = e^{i2\theta}. Substitute this into the expression: ei2θ−1ei2θ+1\frac{e^{i2\theta} - 1}{e^{i2\theta} + 1}. To simplify, multiply the numerator and denominator by e−iθe^{-i\theta}: e−iθ(ei2θ−1)e−iθ(ei2θ+1)=eiθ−e−iθeiθ+e−iθ\frac{e^{-i\theta}(e^{i2\theta} - 1)}{e^{-i\theta}(e^{i2\theta} + 1)} = \frac{e^{i\theta} - e^{-i\theta}}{e^{i\theta} + e^{-i\theta}}. Using Euler's relations, cos⁡θ=eiθ+e−iθ2\cos\theta = \frac{e^{i\theta}+e^{-i\theta}}{2} and sin⁡θ=eiθ−e−iθ2i\sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}, we can rewrite the numerator as 2isin⁡θ2i\sin\theta and the denominator as 2cos⁡θ2\cos\theta. The expression becomes 2isin⁡θ2cos⁡θ=itan⁡θ\frac{2i\sin\theta}{2\cos\theta} = i\tan\theta.

Question 17

The locus of a point zz in the complex plane is given by ∣z−4∣=∣z−2i∣|z - 4| = |z - 2i|. Find the argument of the point on this locus that is closest to the origin.

  1. arctan⁡(−2)\arctan(-2)
  2. arctan⁡(−1/2)\arctan(-1/2) (correct answer)
  3. arctan⁡(1/2)\arctan(1/2)
  4. arctan⁡(2)\arctan(2)
Explanation: The equation ∣z−z1∣=∣z−z2∣|z-z_1| = |z-z_2| describes the perpendicular bisector of the line segment connecting z1z_1 and z2z_2. Here, z1=4z_1 = 4 and z2=2iz_2 = 2i. The segment connects the points (4, 0) and (0, 2). The midpoint of this segment is (4+02,0+22)=(2,1)\left(\frac{4+0}{2}, \frac{0+2}{2}\right) = (2, 1). The slope of the segment is 2−00−4=−12\frac{2-0}{0-4} = -\frac{1}{2}. The slope of the perpendicular bisector is the negative reciprocal, which is 2. The equation of the locus is the line passing through (2, 1) with slope 2: y−1=2(x−2)⇒y=2x−3y-1 = 2(x-2) \Rightarrow y = 2x - 3. The point on this line closest to the origin is the foot of the perpendicular from the origin to the line. The line from the origin perpendicular to the locus has a slope of −12-\frac{1}{2}, so its equation is y=−12xy = -\frac{1}{2}x. To find the intersection point, we set the y-values equal: 2x−3=−12x⇒4x−6=−x⇒5x=6⇒x=652x - 3 = -\frac{1}{2}x \Rightarrow 4x - 6 = -x \Rightarrow 5x = 6 \Rightarrow x = \frac{6}{5}. Then y=−12(65)=−35y = -\frac{1}{2}(\frac{6}{5}) = -\frac{3}{5}. The point is z=65−35iz = \frac{6}{5} - \frac{3}{5}i. The argument is arg⁡(z)=arctan⁡(−3/56/5)=arctan⁡(−12)\arg(z) = \arctan\left(\frac{-3/5}{6/5}\right) = \arctan(-\frac{1}{2}).

Question 18

If z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta, which of the following expressions is equivalent to zn+z−nz^n + z^{-n}?

  1. cos⁡(nθ)\cos(n\theta)
  2. 2cos⁡(nθ)2\cos(n\theta) (correct answer)
  3. 2isin⁡(nθ)2i\sin(n\theta)
  4. 2cos⁡nθ2\cos^n\theta
Explanation: Given z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta, we can use De Moivre's theorem to find znz^n and z−nz^{-n}. According to the theorem, zn=(cos⁡θ+isin⁡θ)n=cos⁡(nθ)+isin⁡(nθ)z^n = (\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta). Similarly, z−n=(cos⁡θ+isin⁡θ)−n=cos⁡(−nθ)+isin⁡(−nθ)z^{-n} = (\cos\theta + i\sin\theta)^{-n} = \cos(-n\theta) + i\sin(-n\theta). Using the properties of trigonometric functions, cos⁡(−x)=cos⁡(x)\cos(-x) = \cos(x) and sin⁡(−x)=−sin⁡(x)\sin(-x) = -\sin(x), we can simplify z−nz^{-n} to cos⁡(nθ)−isin⁡(nθ)\cos(n\theta) - i\sin(n\theta). Now, we add the two expressions: zn+z−n=(cos⁡(nθ)+isin⁡(nθ))+(cos⁡(nθ)−isin⁡(nθ))z^n + z^{-n} = (\cos(n\theta) + i\sin(n\theta)) + (\cos(n\theta) - i\sin(n\theta)). The imaginary parts cancel out, leaving 2cos⁡(nθ)2\cos(n\theta).

Question 19

Which of the following is a solution to the equation ez=3−ie^z = \sqrt{3} - i?

  1. 2−iπ62 - i\frac{\pi}{6}
  2. ln⁡2+iπ6\ln 2 + i\frac{\pi}{6}
  3. ln⁡2−i5π6\ln 2 - i\frac{5\pi}{6}
  4. ln⁡2−iπ6\ln 2 - i\frac{\pi}{6} (correct answer)
Explanation: Let z=x+iyz = x+iy. The equation is ex+iy=3−ie^{x+iy} = \sqrt{3} - i. This can be written as exeiy=3−ie^x e^{iy} = \sqrt{3} - i. We need to write the right side in exponential form, reiθre^{i\theta}. The modulus is r=∣3−i∣=(3)2+(−1)2=3+1=2r = |\sqrt{3} - i| = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3+1} = 2. The argument θ\theta is in the fourth quadrant, so θ=arctan⁡(−13)=−π6\theta = \arctan(\frac{-1}{\sqrt{3}}) = -\frac{\pi}{6}. Thus, 3−i=2e−iπ/6\sqrt{3} - i = 2e^{-i\pi/6}. The equation becomes exeiy=2ei(−π/6+2kπ)e^x e^{iy} = 2e^{i(-\pi/6 + 2k\pi)} for any integer kk. By equating the moduli, we get ex=2e^x = 2, which implies x=ln⁡2x = \ln 2. By equating the arguments, we get y=−π6+2kπy = -\frac{\pi}{6} + 2k\pi. For k=0k=0, we get a solution z=x+iy=ln⁡2−iπ6z = x+iy = \ln 2 - i\frac{\pi}{6}.

Question 20

A complex number zz has modulus rr and argument θ\theta. The complex number w=zzˉ−zˉzw = \frac{z}{\bar{z}} - \frac{\bar{z}}{z} is purely imaginary. What is its imaginary part?

  1. 2sin⁡(2θ)2\sin(2\theta) (correct answer)
  2. 2isin⁡(2θ)2i\sin(2\theta)
  3. 2cos⁡(2θ)2\cos(2\theta)
  4. sin⁡(2θ)\sin(2\theta)
Explanation: Let z=reiθz = re^{i\theta}. Then its conjugate is zˉ=re−iθ\bar{z} = re^{-i\theta}. Now we can evaluate the terms in the expression for ww. zzˉ=reiθre−iθ=eiθ−(−iθ)=ei2θ\frac{z}{\bar{z}} = \frac{re^{i\theta}}{re^{-i\theta}} = e^{i\theta - (-i\theta)} = e^{i2\theta}. zˉz=re−iθreiθ=e−iθ−iθ=e−i2θ\frac{\bar{z}}{z} = \frac{re^{-i\theta}}{re^{i\theta}} = e^{-i\theta - i\theta} = e^{-i2\theta}. So, w=ei2θ−e−i2θw = e^{i2\theta} - e^{-i2\theta}. Using Euler's formula, eiϕ=cos⁡ϕ+isin⁡ϕe^{i\phi} = \cos\phi + i\sin\phi, we have w=(cos⁡(2θ)+isin⁡(2θ))−(cos⁡(−2θ)+isin⁡(−2θ))w = (\cos(2\theta) + i\sin(2\theta)) - (\cos(-2\theta) + i\sin(-2\theta)). This simplifies to w=(cos⁡(2θ)+isin⁡(2θ))−(cos⁡(2θ)−isin⁡(2θ))=2isin⁡(2θ)w = (\cos(2\theta) + i\sin(2\theta)) - (\cos(2\theta) - i\sin(2\theta)) = 2i\sin(2\theta). This is a purely imaginary number. The imaginary part is the coefficient of ii, which is 2sin⁡(2θ)2\sin(2\theta). Note that 2isin⁡(2θ)2i\sin(2\theta) is the complex number itself, not its imaginary part.