IB Mathematics: Analysis and Approaches Quiz: Complex Numbers And Trig
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Complex Numbers And TrigQuestion 1 of 20

Let z1z_1 and z2z_2 be the two roots of the equation z2=2+2i3z^2 = 2+2i\sqrt{3}. Find the value of ∣arg(z1)−arg(z2)∣|\text{arg}(z_1) - \text{arg}(z_2)|, where arguments are the principal arguments in (−π,π](-\pi, \pi].

π6\frac{\pi}{6}
π3\frac{\pi}{3}
2π3\frac{2\pi}{3}
π\pi
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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Complex Numbers And Trig

Practice Complex Numbers And Trig 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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Question 1

Let z1z_1 and z2z_2 be the two roots of the equation z2=2+2i3z^2 = 2+2i\sqrt{3}. Find the value of ∣arg(z1)−arg(z2)∣|\text{arg}(z_1) - \text{arg}(z_2)|, where arguments are the principal arguments in (−π,π](-\pi, \pi].

  1. π6\frac{\pi}{6}
  2. π3\frac{\pi}{3}
  3. 2π3\frac{2\pi}{3}
  4. π\pi (correct answer)
Explanation: Geometrically, the two square roots of a non-zero complex number ww are of the form z1z_1 and z2=−z1z_2 = -z_1. On the Argand diagram, they are separated by an angle of π\pi radians. Therefore, the difference in their arguments will be π\pi. To verify by calculation: First, convert w=2+2i3w = 2+2i\sqrt{3} to polar form. Modulus ∣w∣=22+(23)2=4+12=4|w| = \sqrt{2^2+(2\sqrt{3})^2} = \sqrt{4+12}=4. Argument Arg(w)=arctan⁡(232)=arctan⁡(3)=π3\text{Arg}(w) = \arctan(\frac{2\sqrt{3}}{2}) = \arctan(\sqrt{3}) = \frac{\pi}{3}. So w=4eiπ/3w = 4e^{i\pi/3}. The roots of z2=wz^2 = w are given by zk=4ei(π/3+2kπ)/2z_k = \sqrt{4}e^{i(\pi/3+2k\pi)/2} for k=0,1k=0,1. For k=0k=0: z1=2eiπ/6z_1 = 2e^{i\pi/6}. The argument is arg(z1)=π6\text{arg}(z_1) = \frac{\pi}{6}. For k=1k=1: z2=2ei(π/6+π)=2ei7π/6z_2 = 2e^{i(\pi/6+\pi)} = 2e^{i7\pi/6}. The principal argument is 7π/6−2π=−5π/67\pi/6 - 2\pi = -5\pi/6. The difference is ∣π6−(−5π6)∣=∣6π6∣=π|\frac{\pi}{6} - (-\frac{5\pi}{6})| = |\frac{6\pi}{6}| = \pi. Distractor B is the argument of the original number ww. Distractor A is the argument of one of the roots. Distractor C is double the argument of the original number.

Question 2

Let z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta. The expression for −4sin⁡3θ-4\sin^3\theta can be derived from the binomial expansion of (z−z−1)3(z-z^{-1})^3. What is this expression in terms of sines of multiple angles?

  1. 3sin⁡θ−sin⁡(3θ)3\sin\theta - \sin(3\theta)
  2. sin⁡(3θ)−3sin⁡θ\sin(3\theta) - 3\sin\theta (correct answer)
  3. cos⁡(3θ)−3cos⁡θ\cos(3\theta) - 3\cos\theta
  4. 3cos⁡θ−cos⁡(3θ)3\cos\theta - \cos(3\theta)
Explanation: We use the identity z−z−1=2isin⁡θz - z^{-1} = 2i\sin\theta. Cubing both sides gives (z−z−1)3=(2isin⁡θ)3=8i3sin⁡3θ=−8isin⁡3θ(z-z^{-1})^3 = (2i\sin\theta)^3 = 8i^3\sin^3\theta = -8i\sin^3\theta. We also expand (z−z−1)3(z-z^{-1})^3 binomially: (z−z−1)3=z3−3z2(z−1)+3z(z−1)2−(z−1)3=z3−3z+3z−1−z−3(z-z^{-1})^3 = z^3 - 3z^2(z^{-1}) + 3z(z^{-1})^2 - (z^{-1})^3 = z^3 - 3z + 3z^{-1} - z^{-3}. Grouping terms: (z3−z−3)−3(z−z−1)(z^3-z^{-3}) - 3(z-z^{-1}). Using the identity zn−z−n=2isin⁡(nθ)z^n-z^{-n} = 2i\sin(n\theta), this becomes: 2isin⁡(3θ)−3(2isin⁡θ)=2i(sin⁡(3θ)−3sin⁡θ)2i\sin(3\theta) - 3(2i\sin\theta) = 2i(\sin(3\theta) - 3\sin\theta). Equating the two expressions for (z−z−1)3(z-z^{-1})^3: −8isin⁡3θ=2i(sin⁡(3θ)−3sin⁡θ)-8i\sin^3\theta = 2i(\sin(3\theta) - 3\sin\theta). Dividing both sides by −2i-2i gives: 4sin⁡3θ=−(sin⁡(3θ)−3sin⁡θ)=3sin⁡θ−sin⁡(3θ)4\sin^3\theta = -(\sin(3\theta) - 3\sin\theta) = 3\sin\theta - \sin(3\theta). The question asks for −4sin⁡3θ-4\sin^3\theta, so we multiply by -1: −4sin⁡3θ=−(3sin⁡θ−sin⁡(3θ))=sin⁡(3θ)−3sin⁡θ-4\sin^3\theta = -(3\sin\theta - \sin(3\theta)) = \sin(3\theta) - 3\sin\theta. Distractor A is the expression for +4sin⁡3θ+4\sin^3\theta. Distractors C and D are incorrect expressions involving cosines.

Question 3

Using De Moivre's theorem, an expression for cos⁡(4θ)\cos(4\theta) can be found by considering the real part of (cos⁡θ+isin⁡θ)4(\cos\theta + i\sin\theta)^4. Which of the following is the correct expression in terms of cos⁡θ\cos\theta?

  1. 4cos⁡3θ−3cos⁡θ4\cos^3\theta - 3\cos\theta
  2. 8cos⁡4θ−4cos⁡2θ+18\cos^4\theta - 4\cos^2\theta + 1
  3. 8cos⁡4θ−8cos⁡2θ+18\cos^4\theta - 8\cos^2\theta + 1 (correct answer)
  4. cos⁡4θ−6cos⁡2θsin⁡2θ+sin⁡4θ\cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta
Explanation: By De Moivre's theorem, cos⁡(4θ)+isin⁡(4θ)=(cos⁡θ+isin⁡θ)4\cos(4\theta) + i\sin(4\theta) = (\cos\theta + i\sin\theta)^4. We expand the right side using the binomial theorem: (cos⁡θ+isin⁡θ)4=(40)cos⁡4θ+(41)cos⁡3θ(isin⁡θ)+(42)cos⁡2θ(isin⁡θ)2+(43)cos⁡θ(isin⁡θ)3+(44)(isin⁡θ)4(\cos\theta + i\sin\theta)^4 = \binom{4}{0}\cos^4\theta + \binom{4}{1}\cos^3\theta(i\sin\theta) + \binom{4}{2}\cos^2\theta(i\sin\theta)^2 + \binom{4}{3}\cos\theta(i\sin\theta)^3 + \binom{4}{4}(i\sin\theta)^4 =cos⁡4θ+4icos⁡3θsin⁡θ−6cos⁡2θsin⁡2θ−4icos⁡θsin⁡3θ+sin⁡4θ= \cos^4\theta + 4i\cos^3\theta\sin\theta - 6\cos^2\theta\sin^2\theta - 4i\cos\theta\sin^3\theta + \sin^4\theta The real part is cos⁡(4θ)=cos⁡4θ−6cos⁡2θsin⁡2θ+sin⁡4θ\cos(4\theta) = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta. To express this in terms of cos⁡θ\cos\theta, we substitute sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta: cos⁡(4θ)=cos⁡4θ−6cos⁡2θ(1−cos⁡2θ)+(1−cos⁡2θ)2\cos(4\theta) = \cos^4\theta - 6\cos^2\theta(1 - \cos^2\theta) + (1 - \cos^2\theta)^2 =cos⁡4θ−6cos⁡2θ+6cos⁡4θ+(1−2cos⁡2θ+cos⁡4θ)= \cos^4\theta - 6\cos^2\theta + 6\cos^4\theta + (1 - 2\cos^2\theta + \cos^4\theta) =(1+6+1)cos⁡4θ+(−6−2)cos⁡2θ+1= (1+6+1)\cos^4\theta + (-6-2)\cos^2\theta + 1 =8cos⁡4θ−8cos⁡2θ+1= 8\cos^4\theta - 8\cos^2\theta + 1. Distractor A is the expression for cos⁡(3θ)\cos(3\theta). Distractor B has an algebraic error in collecting the cos⁡2θ\cos^2\theta terms. Distractor D is the correct real part before converting sine terms to cosine terms.

Question 4

Let z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta. The expression for −4sin⁡3θ-4\sin^3\theta can be derived from the binomial expansion of (z−z−1)3(z-z^{-1})^3. What is this expression in terms of sines of multiple angles?

  1. 3sin⁡θ−sin⁡(3θ)3\sin\theta - \sin(3\theta)
  2. sin⁡(3θ)−3sin⁡θ\sin(3\theta) - 3\sin\theta (correct answer)
  3. cos⁡(3θ)−3cos⁡θ\cos(3\theta) - 3\cos\theta
  4. 3cos⁡θ−cos⁡(3θ)3\cos\theta - \cos(3\theta)
Explanation: We use the identity z−z−1=2isin⁡θz - z^{-1} = 2i\sin\theta. Cubing both sides gives (z−z−1)3=(2isin⁡θ)3=8i3sin⁡3θ=−8isin⁡3θ(z-z^{-1})^3 = (2i\sin\theta)^3 = 8i^3\sin^3\theta = -8i\sin^3\theta. We also expand (z−z−1)3(z-z^{-1})^3 binomially: (z−z−1)3=z3−3z2(z−1)+3z(z−1)2−(z−1)3=z3−3z+3z−1−z−3(z-z^{-1})^3 = z^3 - 3z^2(z^{-1}) + 3z(z^{-1})^2 - (z^{-1})^3 = z^3 - 3z + 3z^{-1} - z^{-3}. Grouping terms: (z3−z−3)−3(z−z−1)(z^3-z^{-3}) - 3(z-z^{-1}). Using the identity zn−z−n=2isin⁡(nθ)z^n-z^{-n} = 2i\sin(n\theta), this becomes: 2isin⁡(3θ)−3(2isin⁡θ)=2i(sin⁡(3θ)−3sin⁡θ)2i\sin(3\theta) - 3(2i\sin\theta) = 2i(\sin(3\theta) - 3\sin\theta). Equating the two expressions for (z−z−1)3(z-z^{-1})^3: −8isin⁡3θ=2i(sin⁡(3θ)−3sin⁡θ)-8i\sin^3\theta = 2i(\sin(3\theta) - 3\sin\theta). Dividing both sides by −2i-2i gives: 4sin⁡3θ=−(sin⁡(3θ)−3sin⁡θ)=3sin⁡θ−sin⁡(3θ)4\sin^3\theta = -(\sin(3\theta) - 3\sin\theta) = 3\sin\theta - \sin(3\theta). The question asks for −4sin⁡3θ-4\sin^3\theta, so we multiply by -1: −4sin⁡3θ=−(3sin⁡θ−sin⁡(3θ))=sin⁡(3θ)−3sin⁡θ-4\sin^3\theta = -(3\sin\theta - \sin(3\theta)) = \sin(3\theta) - 3\sin\theta. Distractor A is the expression for +4sin⁡3θ+4\sin^3\theta. Distractors C and D are incorrect expressions involving cosines.

Question 5

Let z1z_1 and z2z_2 be the two roots of the equation z2=2+2i3z^2 = 2+2i\sqrt{3}. Find the value of ∣arg(z1)−arg(z2)∣|\text{arg}(z_1) - \text{arg}(z_2)|, where arguments are the principal arguments in (−π,π](-\pi, \pi].

  1. π6\frac{\pi}{6}
  2. π3\frac{\pi}{3}
  3. 2π3\frac{2\pi}{3}
  4. π\pi (correct answer)
Explanation: Geometrically, the two square roots of a non-zero complex number ww are of the form z1z_1 and z2=−z1z_2 = -z_1. On the Argand diagram, they are separated by an angle of π\pi radians. Therefore, the difference in their arguments will be π\pi. To verify by calculation: First, convert w=2+2i3w = 2+2i\sqrt{3} to polar form. Modulus ∣w∣=22+(23)2=4+12=4|w| = \sqrt{2^2+(2\sqrt{3})^2} = \sqrt{4+12}=4. Argument Arg(w)=arctan⁡(232)=arctan⁡(3)=π3\text{Arg}(w) = \arctan(\frac{2\sqrt{3}}{2}) = \arctan(\sqrt{3}) = \frac{\pi}{3}. So w=4eiπ/3w = 4e^{i\pi/3}. The roots of z2=wz^2 = w are given by zk=4ei(π/3+2kπ)/2z_k = \sqrt{4}e^{i(\pi/3+2k\pi)/2} for k=0,1k=0,1. For k=0k=0: z1=2eiπ/6z_1 = 2e^{i\pi/6}. The argument is arg(z1)=π6\text{arg}(z_1) = \frac{\pi}{6}. For k=1k=1: z2=2ei(π/6+π)=2ei7π/6z_2 = 2e^{i(\pi/6+\pi)} = 2e^{i7\pi/6}. The principal argument is 7π/6−2π=−5π/67\pi/6 - 2\pi = -5\pi/6. The difference is ∣π6−(−5π6)∣=∣6π6∣=π|\frac{\pi}{6} - (-\frac{5\pi}{6})| = |\frac{6\pi}{6}| = \pi. Distractor B is the argument of the original number ww. Distractor A is the argument of one of the roots. Distractor C is double the argument of the original number.

Question 6

Using De Moivre's theorem, an expression for cos⁡(4θ)\cos(4\theta) can be found by considering the real part of (cos⁡θ+isin⁡θ)4(\cos\theta + i\sin\theta)^4. Which of the following is the correct expression in terms of cos⁡θ\cos\theta?

  1. 4cos⁡3θ−3cos⁡θ4\cos^3\theta - 3\cos\theta
  2. 8cos⁡4θ−4cos⁡2θ+18\cos^4\theta - 4\cos^2\theta + 1
  3. 8cos⁡4θ−8cos⁡2θ+18\cos^4\theta - 8\cos^2\theta + 1 (correct answer)
  4. cos⁡4θ−6cos⁡2θsin⁡2θ+sin⁡4θ\cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta
Explanation: By De Moivre's theorem, cos⁡(4θ)+isin⁡(4θ)=(cos⁡θ+isin⁡θ)4\cos(4\theta) + i\sin(4\theta) = (\cos\theta + i\sin\theta)^4. We expand the right side using the binomial theorem: (cos⁡θ+isin⁡θ)4=(40)cos⁡4θ+(41)cos⁡3θ(isin⁡θ)+(42)cos⁡2θ(isin⁡θ)2+(43)cos⁡θ(isin⁡θ)3+(44)(isin⁡θ)4(\cos\theta + i\sin\theta)^4 = \binom{4}{0}\cos^4\theta + \binom{4}{1}\cos^3\theta(i\sin\theta) + \binom{4}{2}\cos^2\theta(i\sin\theta)^2 + \binom{4}{3}\cos\theta(i\sin\theta)^3 + \binom{4}{4}(i\sin\theta)^4 =cos⁡4θ+4icos⁡3θsin⁡θ−6cos⁡2θsin⁡2θ−4icos⁡θsin⁡3θ+sin⁡4θ= \cos^4\theta + 4i\cos^3\theta\sin\theta - 6\cos^2\theta\sin^2\theta - 4i\cos\theta\sin^3\theta + \sin^4\theta The real part is cos⁡(4θ)=cos⁡4θ−6cos⁡2θsin⁡2θ+sin⁡4θ\cos(4\theta) = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta. To express this in terms of cos⁡θ\cos\theta, we substitute sin⁡2θ=1−cos⁡2θ\sin^2\theta = 1 - \cos^2\theta: cos⁡(4θ)=cos⁡4θ−6cos⁡2θ(1−cos⁡2θ)+(1−cos⁡2θ)2\cos(4\theta) = \cos^4\theta - 6\cos^2\theta(1 - \cos^2\theta) + (1 - \cos^2\theta)^2 =cos⁡4θ−6cos⁡2θ+6cos⁡4θ+(1−2cos⁡2θ+cos⁡4θ)= \cos^4\theta - 6\cos^2\theta + 6\cos^4\theta + (1 - 2\cos^2\theta + \cos^4\theta) =(1+6+1)cos⁡4θ+(−6−2)cos⁡2θ+1= (1+6+1)\cos^4\theta + (-6-2)\cos^2\theta + 1 =8cos⁡4θ−8cos⁡2θ+1= 8\cos^4\theta - 8\cos^2\theta + 1. Distractor A is the expression for cos⁡(3θ)\cos(3\theta). Distractor B has an algebraic error in collecting the cos⁡2θ\cos^2\theta terms. Distractor D is the correct real part before converting sine terms to cosine terms.

Question 7

If z=eiθz = e^{i\theta}, what is the modulus of 1+z+z21+z+z^2?

  1. 3
  2. ∣cos⁡(3θ/2)cos⁡(θ/2)∣\left| \frac{\cos(3\theta/2)}{\cos(\theta/2)} \right|
  3. ∣sin⁡(3θ)sin⁡(θ)∣\left| \frac{\sin(3\theta)}{\sin(\theta)} \right|
  4. ∣sin⁡(3θ/2)sin⁡(θ/2)∣\left| \frac{\sin(3\theta/2)}{\sin(\theta/2)} \right| (correct answer)
Explanation: We can express 1+z+z21+z+z^2 as the sum of a geometric series: 1+z+z2=z3−1z−11+z+z^2 = \frac{z^3-1}{z-1}. Therefore, ∣1+z+z2∣=∣z3−1∣∣z−1∣|1+z+z^2| = \frac{|z^3-1|}{|z-1|}. Let's evaluate ∣zn−1∣|z^n-1| where z=eiθz=e^{i\theta}. ∣zn−1∣2=(zn−1)(zn‾−1)=(zn−1)(z−n−1)=znz−n−zn−z−n+1=1−(zn+z−n)+1=2−2cos⁡(nθ)|z^n-1|^2 = (z^n-1)(\overline{z^n}-1) = (z^n-1)(z^{-n}-1) = z^n z^{-n} - z^n - z^{-n} + 1 = 1 - (z^n+z^{-n}) + 1 = 2 - 2\cos(n\theta). Using the half-angle identity 1−cos⁡(A)=2sin⁡2(A/2)1-\cos(A) = 2\sin^2(A/2), this becomes 2(1−cos⁡(nθ))=4sin⁡2(nθ/2)2(1-\cos(n\theta)) = 4\sin^2(n\theta/2). So, ∣zn−1∣=2∣sin⁡(nθ/2)∣|z^n-1| = 2|\sin(n\theta/2)|. Applying this to our expression: ∣1+z+z2∣=2∣sin⁡(3θ/2)∣2∣sin⁡(θ/2)∣=∣sin⁡(3θ/2)sin⁡(θ/2)∣|1+z+z^2| = \frac{2|\sin(3\theta/2)|}{2|\sin(\theta/2)|} = \left| \frac{\sin(3\theta/2)}{\sin(\theta/2)} \right|. Distractor A is the result of incorrectly applying the triangle inequality: ∣1+z+z2∣≤∣1∣+∣z∣+∣z2∣=1+1+1=3|1+z+z^2| \leq |1|+|z|+|z^2| = 1+1+1=3. Distractor B uses the wrong half-angle identity (for 1+cos⁡A1+\cos A). Distractor C makes an error in the half-angle formula application.

Question 8

If z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta, which of the following real expressions is equivalent to (z2+1)2z2\frac{(z^2+1)^2}{z^2}?

  1. 2cos⁡(2θ)+22\cos(2\theta)+2
  2. 4cos⁡2θ4\cos^2\theta (correct answer)
  3. 4sin⁡2θ4\sin^2\theta
  4. 2cos⁡2(2θ)2\cos^2(2\theta)
Explanation: We can rewrite the expression by dividing the numerator by the denominator: (z2+1)2z2=z4+2z2+1z2=z2+2+z−2\frac{(z^2+1)^2}{z^2} = \frac{z^4+2z^2+1}{z^2} = z^2 + 2 + z^{-2}. Now, we group the terms: (z2+z−2)+2(z^2 + z^{-2}) + 2. Using the identity zn+z−n=2cos⁡(nθ)z^n + z^{-n} = 2\cos(n\theta), we have: z2+z−2=2cos⁡(2θ)z^2 + z^{-2} = 2\cos(2\theta). So the expression becomes 2cos⁡(2θ)+22\cos(2\theta) + 2. Using the double angle identity for cosine, cos⁡(2θ)=2cos⁡2θ−1\cos(2\theta) = 2\cos^2\theta - 1, we can simplify further: 2(2cos⁡2θ−1)+2=4cos⁡2θ−2+2=4cos⁡2θ2(2\cos^2\theta - 1) + 2 = 4\cos^2\theta - 2 + 2 = 4\cos^2\theta. Distractor A is a correct intermediate step, but not the most simplified form. Distractor C may result from a sign error or using the wrong double angle identity. Distractor D is an incorrect simplification.

Question 9

Given z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta, the expression z5−z−5z−z−1\frac{z^5 - z^{-5}}{z - z^{-1}} can be written as a sum of cosines. What is this sum?

  1. 1+2cos⁡(2θ)+2cos⁡(4θ)1 + 2\cos(2\theta) + 2\cos(4\theta) (correct answer)
  2. 2cos⁡(2θ)+2cos⁡(4θ)2\cos(2\theta) + 2\cos(4\theta)
  3. 1+cos⁡(2θ)+cos⁡(4θ)1 + \cos(2\theta) + \cos(4\theta)
  4. sin⁡(5θ)sin⁡θ\frac{\sin(5\theta)}{\sin\theta}
Explanation: Let the expression be EE. We can recognize z5−z−5z^5 - z^{-5} as z−5(z10−1)z^{-5}(z^{10}-1) and z−z−1z - z^{-1} as z−1(z2−1)z^{-1}(z^2-1). E=z−5(z10−1)z−1(z2−1)=z−4(z2−1)(z8+z6+z4+z2+1)z2−1=z−4(z8+z6+z4+z2+1)E = \frac{z^{-5}(z^{10}-1)}{z^{-1}(z^2-1)} = z^{-4}\frac{(z^2-1)(z^8+z^6+z^4+z^2+1)}{z^2-1} = z^{-4}(z^8+z^6+z^4+z^2+1). Distributing z−4z^{-4} gives: E=z4+z2+1+z−2+z−4E = z^4 + z^2 + 1 + z^{-2} + z^{-4}. Now, we group the terms: E=1+(z2+z−2)+(z4+z−4)E = 1 + (z^2 + z^{-2}) + (z^4 + z^{-4}). Using the identity zn+z−n=2cos⁡(nθ)z^n + z^{-n} = 2\cos(n\theta), we have: E=1+2cos⁡(2θ)+2cos⁡(4θ)E = 1 + 2\cos(2\theta) + 2\cos(4\theta). Distractor B is missing the constant term '1'. Distractor C is missing the factors of 2. Distractor D is an alternative, equivalent form of the expression using the identity zn−z−n=2isin⁡(nθ)z^n - z^{-n} = 2i\sin(n\theta), which leads to 2isin⁡(5θ)2isin⁡θ\frac{2i\sin(5\theta)}{2i\sin\theta}, but the question asks for a sum of cosines.

Question 10

Let z=12−32iz = \frac{1}{2} - \frac{\sqrt{3}}{2}i. Find the value of z2023z^{2023}.

  1. -1
  2. 1
  3. 12−32i\frac{1}{2} - \frac{\sqrt{3}}{2}i (correct answer)
  4. 12+32i\frac{1}{2} + \frac{\sqrt{3}}{2}i
Explanation: First, convert zz to its polar form, r(cos⁡θ+isin⁡θ)r(\cos\theta + i\sin\theta) or reiθre^{i\theta}. The modulus is r=∣z∣=(12)2+(−32)2=14+34=1r = |z| = \sqrt{(\frac{1}{2})^2 + (-\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1. The argument is θ=arctan⁡(−3/21/2)=arctan⁡(−3)=−π3\theta = \arctan\left(\frac{-\sqrt{3}/2}{1/2}\right) = \arctan(-\sqrt{3}) = -\frac{\pi}{3} (since zz is in the fourth quadrant). So, z=1⋅e−iπ/3=e−iπ/3z = 1 \cdot e^{-i\pi/3} = e^{-i\pi/3}. Using De Moivre's theorem, z2023=(e−iπ/3)2023=e−i2023π/3z^{2023} = (e^{-i\pi/3})^{2023} = e^{-i2023\pi/3}. To simplify the argument, we find the remainder of 2023 divided by 6 (the period of eikπ/3e^{ik\pi/3}). Or we can write −2023π/3-2023\pi/3 as (−674⋅3−1)π/3=−674π−π/3(-674 \cdot 3 - 1)\pi/3 = -674\pi - \pi/3. Since −674π-674\pi is an integer multiple of 2π2\pi, the angle is coterminal with −π/3-\pi/3. So, z2023=e−iπ/3=z=12−32iz^{2023} = e^{-i\pi/3} = z = \frac{1}{2} - \frac{\sqrt{3}}{2}i. Distractor A and B are common incorrect results from miscalculating the argument's periodicity. Distractor D results from a sign error in the original argument of zz.

Question 11

Let ω\omega be a non-real cube root of unity. What is the value of (1+ω−ω2)5(1 + \omega - \omega^2)^5?

  1. −32-32
  2. −32ω-32\omega (correct answer)
  3. −32ω2-32\omega^2
  4. 32ω32\omega
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 satisfy ω2+ω+1=0\omega^2 + \omega + 1 = 0. From this identity, we can write 1+ω=−ω21 + \omega = -\omega^2. Substitute this into the expression: (1+ω−ω2)5=(−ω2−ω2)5=(−2ω2)5(1 + \omega - \omega^2)^5 = (-\omega^2 - \omega^2)^5 = (-2\omega^2)^5. Now, we simplify this expression: (−2ω2)5=(−2)5(ω2)5=−32ω10(-2\omega^2)^5 = (-2)^5 (\omega^2)^5 = -32 \omega^{10}. Since ω\omega is a cube root of unity, ω3=1\omega^3 = 1. We can simplify ω10\omega^{10} as ω10=(ω3)3ω=13ω=ω\omega^{10} = (\omega^3)^3 \omega = 1^3 \omega = \omega. Therefore, the final value is −32ω-32\omega. Distractor A results from incorrectly simplifying ω10\omega^{10} to 1. Distractor C may result from an error in simplifying ω10\omega^{10} or confusion with the other non-real root. Distractor D has a sign error in (−2)5(-2)^5.

Question 12

Consider the equation (z+1)4=z4(z+1)^4 = z^4. Which of the following statements is true about its roots in the complex plane?

  1. The sum of the roots is 0.
  2. The product of the roots is 14\frac{1}{4}.
  3. All roots lie on the line Re(z)=−12\text{Re}(z) = -\frac{1}{2}. (correct answer)
  4. All roots are real.
Explanation: The equation can be rewritten as (z+1z)4=1\left(\frac{z+1}{z}\right)^4 = 1. Let w=z+1zw = \frac{z+1}{z}. The equation becomes w4=1w^4 = 1. The solutions for ww are the 4th roots of unity: 1,i,−1,−i1, i, -1, -i. We have w=1+1zw = 1 + \frac{1}{z}. The case w=1w=1 implies 1=1+1z1 = 1 + \frac{1}{z}, which means 1z=0\frac{1}{z} = 0, an impossibility. So there are only three roots for zz, corresponding to w=i,−1,−iw = i, -1, -i. We solve for zz using z=1w−1z = \frac{1}{w-1}: If w=iw = i: z=1i−1=−1−i2=−12−12iz = \frac{1}{i-1} = \frac{-1-i}{2} = -\frac{1}{2} - \frac{1}{2}i. If w=−1w = -1: z=1−1−1=−12z = \frac{1}{-1-1} = -\frac{1}{2}. If w=−iw = -i: z=1−i−1=−1+i2=−12+12iz = \frac{1}{-i-1} = \frac{-1+i}{2} = -\frac{1}{2} + \frac{1}{2}i. All three roots have a real part of −12-\frac{1}{2}, so they lie on the vertical line Re(z)=−12\text{Re}(z) = -\frac{1}{2}. Alternatively, expanding (z+1)4=z4(z+1)^4=z^4 gives 4z3+6z2+4z+1=04z^3+6z^2+4z+1=0. The sum of roots is −64=−32-\frac{6}{4} = -\frac{3}{2} (A is false). The product of roots is −14-\frac{1}{4} (B is false). D is false as two roots are complex.

Question 13

Let ω\omega be a non-real 7th root of unity. What is the value of the sum S=∑k=06(1+ωk)2S = \sum_{k=0}^{6} (1+\omega^k)^2?

  1. 0
  2. 7 (correct answer)
  3. 14
  4. 21
Explanation: We expand the term inside the summation first: S=∑k=06(1+2ωk+ω2k)S = \sum_{k=0}^{6} (1 + 2\omega^k + \omega^{2k}). We can split the summation into three parts: S=∑k=061+∑k=062ωk+∑k=06ω2kS = \sum_{k=0}^{6} 1 + \sum_{k=0}^{6} 2\omega^k + \sum_{k=0}^{6} \omega^{2k} S=∑k=061+2∑k=06ωk+∑k=06(ω2)kS = \sum_{k=0}^{6} 1 + 2\sum_{k=0}^{6} \omega^k + \sum_{k=0}^{6} (\omega^2)^k. For the nn-th roots of unity, we know that ∑k=0n−1ωk=0\sum_{k=0}^{n-1} \omega^k = 0. Here, n=7n=7, so ∑k=06ωk=0\sum_{k=0}^{6} \omega^k = 0. The first term is ∑k=061=7\sum_{k=0}^{6} 1 = 7. The second term is 2∑k=06ωk=2(0)=02\sum_{k=0}^{6} \omega^k = 2(0) = 0. For the third term, let u=ω2u = \omega^2. Since ω\omega is a non-real 7th root of unity, ω7=1\omega^7=1 and ω≠1\omega \neq 1. Then u7=(ω2)7=(ω7)2=12=1u^7 = (\omega^2)^7 = (\omega^7)^2 = 1^2 = 1. Also, u≠1u \neq 1 since ω2\omega^2 is also a non-real 7th root of unity. So the sum ∑k=06uk=∑k=06(ω2)k=0\sum_{k=0}^{6} u^k = \sum_{k=0}^{6} (\omega^2)^k = 0. Therefore, S=7+0+0=7S = 7 + 0 + 0 = 7. Distractor A incorrectly assumes the entire sum is zero. Distractor C comes from perhaps thinking the third sum is 7, so 7+0+7=147+0+7=14. Distractor D might come from 7+7+7=217+7+7=21.

Question 14

Let z=(cos⁡α+isin⁡α)(cos⁡α−isin⁡α)−3z = (\cos\alpha + i\sin\alpha)(\cos\alpha - i\sin\alpha)^{-3}. Which of the following is equal to zz?

  1. cos⁡(4α)+isin⁡(4α)\cos(4\alpha) + i\sin(4\alpha) (correct answer)
  2. cos⁡(4α)−isin⁡(4α)\cos(4\alpha) - i\sin(4\alpha)
  3. cos⁡(2α)+isin⁡(2α)\cos(2\alpha) + i\sin(2\alpha)
  4. cos⁡(2α)−isin⁡(2α)\cos(2\alpha) - i\sin(2\alpha)
Explanation: We can write the expression in Euler's form. cos⁡α+isin⁡α=eiα\cos\alpha + i\sin\alpha = e^{i\alpha} and cos⁡α−isin⁡α=e−iα\cos\alpha - i\sin\alpha = e^{-i\alpha}. The expression becomes: z=(eiα)(e−iα)−3z = (e^{i\alpha}) (e^{-i\alpha})^{-3} Using the laws of exponents: z=eiα⋅e(−iα)(−3)=eiα⋅ei3α=eiα+i3α=ei4αz = e^{i\alpha} \cdot e^{(-i\alpha)(-3)} = e^{i\alpha} \cdot e^{i3\alpha} = e^{i\alpha + i3\alpha} = e^{i4\alpha}. Converting this back to trigonometric form gives: z=cos⁡(4α)+isin⁡(4α)z = \cos(4\alpha) + i\sin(4\alpha). Distractor B results from a sign error in the final exponent. Distractors C and D result from incorrectly combining the exponents, for example by doing 3−1=23-1=2 instead of 1−(−3)=41 - (-3) = 4.

Question 15

The roots of the equation (z−(1+i))5=32(z - (1+i))^5 = 32 are plotted on the Argand diagram. Which statement accurately describes their geometric arrangement?

  1. The roots lie on a circle of radius 2 centered at the origin.
  2. The sum of the roots is 5+5i5+5i. (correct answer)
  3. The product of the roots is 32.
  4. One of the roots is a real number.
Explanation: Let w=z−(1+i)w = z - (1+i). The equation becomes w5=32w^5 = 32. The solutions for ww are the 5th roots of 32, given by wk=2ei(2kπ/5)w_k = 2e^{i(2k\pi/5)} for k=0,1,2,3,4k=0,1,2,3,4. The sum of these roots of unity (scaled by 2) is ∑k=04wk=0\sum_{k=0}^{4} w_k = 0. The roots for zz are given by zk=wk+(1+i)z_k = w_k + (1+i). This means the set of roots for zz is a translation of the set of roots for ww by the vector corresponding to 1+i1+i. The sum of the roots for zz is: ∑k=04zk=∑k=04(wk+(1+i))=(∑k=04wk)+∑k=04(1+i)=0+5(1+i)=5+5i\sum_{k=0}^{4} z_k = \sum_{k=0}^{4} (w_k + (1+i)) = (\sum_{k=0}^{4} w_k) + \sum_{k=0}^{4} (1+i) = 0 + 5(1+i) = 5+5i. So, statement B is correct. Statement A is false: The roots form a regular pentagon centered at 1+i1+i, not the origin. They lie on a circle of radius 2 centered at 1+i1+i. Statement C is false: Expanding the equation leads to a polynomial whose constant term is −(1+i)5−32-(1+i)^5-32. The product of the roots is complex. Statement D is false: For a root zkz_k to be real, its imaginary part must be zero. zk=(1+2cos⁡(2kπ/5))+i(1+2sin⁡(2kπ/5))z_k = (1+2\cos(2k\pi/5)) + i(1+2\sin(2k\pi/5)). For the imaginary part to be zero, sin⁡(2kπ/5)=−1/2\sin(2k\pi/5) = -1/2. This does not occur for k=0,1,2,3,4k=0,1,2,3,4.

Question 16

Given that zn+z−n=2cos⁡(nθ)z^n + z^{-n} = 2\cos(n\theta), which of the following is equivalent to 16cos⁡5θ16\cos^5\theta?

  1. 10cos⁡θ+5cos⁡(3θ)+cos⁡(5θ)10\cos\theta + 5\cos(3\theta) + \cos(5\theta) (correct answer)
  2. 10sin⁡θ+5sin⁡(3θ)+sin⁡(5θ)10\sin\theta + 5\sin(3\theta) + \sin(5\theta)
  3. 2cos⁡(5θ)+10cos⁡(3θ)+20cos⁡θ2\cos(5\theta) + 10\cos(3\theta) + 20\cos\theta
  4. 2cos⁡(5θ)+5cos⁡(3θ)+10cos⁡θ2\cos(5\theta) + 5\cos(3\theta) + 10\cos\theta
Explanation: We start with 2cos⁡θ=z+z−12\cos\theta = z + z^{-1} and raise both sides to the power of 5. (2cos⁡θ)5=32cos⁡5θ=(z+z−1)5(2\cos\theta)^5 = 32\cos^5\theta = (z + z^{-1})^5. Using the binomial expansion for (z+z−1)5(z+z^{-1})^5, with coefficients 1, 5, 10, 10, 5, 1: z5+5z4(z−1)+10z3(z−1)2+10z2(z−1)3+5z(z−1)4+z−5z^5 + 5z^4(z^{-1}) + 10z^3(z^{-1})^2 + 10z^2(z^{-1})^3 + 5z(z^{-1})^4 + z^{-5} =z5+5z3+10z+10z−1+5z−3+z−5= z^5 + 5z^3 + 10z + 10z^{-1} + 5z^{-3} + z^{-5} Group the terms: =(z5+z−5)+5(z3+z−3)+10(z+z−1)= (z^5 + z^{-5}) + 5(z^3 + z^{-3}) + 10(z + z^{-1}) Substitute the identity zn+z−n=2cos⁡(nθ)z^n + z^{-n} = 2\cos(n\theta): 32cos⁡5θ=2cos⁡(5θ)+5(2cos⁡(3θ))+10(2cos⁡θ)32\cos^5\theta = 2\cos(5\theta) + 5(2\cos(3\theta)) + 10(2\cos\theta) 32cos⁡5θ=2cos⁡(5θ)+10cos⁡(3θ)+20cos⁡θ32\cos^5\theta = 2\cos(5\theta) + 10\cos(3\theta) + 20\cos\theta To find 16cos⁡5θ16\cos^5\theta, we divide by 2: 16cos⁡5θ=cos⁡(5θ)+5cos⁡(3θ)+10cos⁡θ16\cos^5\theta = \cos(5\theta) + 5\cos(3\theta) + 10\cos\theta. Distractor D has the coefficients mixed up. Distractor C is the expression for 32cos⁡5θ32\cos^5\theta but with an error. Distractor B uses sine functions instead of cosine functions.

Question 17

The solutions to the equation z5=−32z^5 = -32 form the vertices of a regular pentagon in the Argand diagram. What is the area of this pentagon?

  1. 10sin⁡(π5)10\sin(\frac{\pi}{5})
  2. 10sin⁡(2π5)10\sin(\frac{2\pi}{5}) (correct answer)
  3. 20sin⁡(π5)20\sin(\frac{\pi}{5})
  4. 20sin⁡(2π5)20\sin(\frac{2\pi}{5})
Explanation: The equation is z5=−32z^5 = -32. In polar form, −32=32eiπ-32 = 32e^{i\pi}. The solutions are zk=(32ei(π+2kπ))1/5=2ei(π+2kπ)/5z_k = (32e^{i(\pi+2k\pi)})^{1/5} = 2e^{i(\pi+2k\pi)/5} for k=0,1,2,3,4k=0,1,2,3,4. All five roots have a modulus of r=2r=2, so they lie on a circle of radius 2 centered at the origin. These roots form the vertices of a regular pentagon. The pentagon can be divided into 5 congruent isosceles triangles with two sides of length r=2r=2 and the angle between them being 2π5\frac{2\pi}{5}. The area of one such triangle is 12absin⁡C=12(2)(2)sin⁡(2π5)=2sin⁡(2π5)\frac{1}{2}ab\sin C = \frac{1}{2}(2)(2)\sin(\frac{2\pi}{5}) = 2\sin(\frac{2\pi}{5}). The total area of the pentagon is 5 times the area of one triangle: Area =5×2sin⁡(2π5)=10sin⁡(2π5)= 5 \times 2\sin(\frac{2\pi}{5}) = 10\sin(\frac{2\pi}{5}). Distractor A uses an incorrect central angle of π/5\pi/5. Distractor D forgets the 1/21/2 in the triangle area formula. Distractor C makes both of these errors.

Question 18

Let z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta. Using the identity 2cos⁡θ=z+z−12\cos\theta = z + z^{-1}, find the expression for 8cos⁡3θ8\cos^3\theta in terms of multiple angles.

  1. cos⁡(3θ)+3cos⁡θ\cos(3\theta) + 3\cos\theta
  2. 2cos⁡(3θ)+6cos⁡θ2\cos(3\theta) + 6\cos\theta (correct answer)
  3. 2cos⁡(3θ)−6cos⁡θ2\cos(3\theta) - 6\cos\theta
  4. 2sin⁡(3θ)+6sin⁡θ2\sin(3\theta) + 6\sin\theta
Explanation: We start with 2cos⁡θ=z+z−12\cos\theta = z + z^{-1}. Cubing both sides gives: (2cos⁡θ)3=(z+z−1)3(2\cos\theta)^3 = (z + z^{-1})^3 8cos⁡3θ=z3+3z2(z−1)+3z(z−1)2+(z−1)38\cos^3\theta = z^3 + 3z^2(z^{-1}) + 3z(z^{-1})^2 + (z^{-1})^3 8cos⁡3θ=z3+3z+3z−1+z−38\cos^3\theta = z^3 + 3z + 3z^{-1} + z^{-3} Group the terms: 8cos⁡3θ=(z3+z−3)+3(z+z−1)8\cos^3\theta = (z^3 + z^{-3}) + 3(z + z^{-1}). Using the general identity zn+z−n=2cos⁡(nθ)z^n + z^{-n} = 2\cos(n\theta), we substitute: 8cos⁡3θ=(2cos⁡(3θ))+3(2cos⁡θ)8\cos^3\theta = (2\cos(3\theta)) + 3(2\cos\theta) 8cos⁡3θ=2cos⁡(3θ)+6cos⁡θ8\cos^3\theta = 2\cos(3\theta) + 6\cos\theta. Distractor A arises from forgetting the factors of 2 in the identities zn+z−n=2cos⁡(nθ)z^n+z^{-n}=2\cos(n\theta). Distractor C would result from expanding (z−z−1)3(z-z^{-1})^3 but incorrectly matching it to cosine terms. Distractor D results from incorrectly using sine identities.

Question 19

Let ω\omega be a non-real cube root of unity. What is the value of (1+ω−ω2)5(1 + \omega - \omega^2)^5?

  1. −32-32
  2. −32ω-32\omega (correct answer)
  3. −32ω2-32\omega^2
  4. 32ω32\omega
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 satisfy ω2+ω+1=0\omega^2 + \omega + 1 = 0. From this identity, we can write 1+ω=−ω21 + \omega = -\omega^2. Substitute this into the expression: (1+ω−ω2)5=(−ω2−ω2)5=(−2ω2)5(1 + \omega - \omega^2)^5 = (-\omega^2 - \omega^2)^5 = (-2\omega^2)^5. Now, we simplify this expression: (−2ω2)5=(−2)5(ω2)5=−32ω10(-2\omega^2)^5 = (-2)^5 (\omega^2)^5 = -32 \omega^{10}. Since ω\omega is a cube root of unity, ω3=1\omega^3 = 1. We can simplify ω10\omega^{10} as ω10=(ω3)3ω=13ω=ω\omega^{10} = (\omega^3)^3 \omega = 1^3 \omega = \omega. Therefore, the final value is −32ω-32\omega. Distractor A results from incorrectly simplifying ω10\omega^{10} to 1. Distractor C may result from an error in simplifying ω10\omega^{10} or confusion with the other non-real root. Distractor D has a sign error in (−2)5(-2)^5.

Question 20

Given z=cos⁡θ+isin⁡θz = \cos\theta + i\sin\theta, the expression z5−z−5z−z−1\frac{z^5 - z^{-5}}{z - z^{-1}} can be written as a sum of cosines. What is this sum?

  1. 1+2cos⁡(2θ)+2cos⁡(4θ)1 + 2\cos(2\theta) + 2\cos(4\theta) (correct answer)
  2. 2cos⁡(2θ)+2cos⁡(4θ)2\cos(2\theta) + 2\cos(4\theta)
  3. 1+cos⁡(2θ)+cos⁡(4θ)1 + \cos(2\theta) + \cos(4\theta)
  4. sin⁡(5θ)sin⁡θ\frac{\sin(5\theta)}{\sin\theta}
Explanation: Let the expression be EE. We can recognize z5−z−5z^5 - z^{-5} as z−5(z10−1)z^{-5}(z^{10}-1) and z−z−1z - z^{-1} as z−1(z2−1)z^{-1}(z^2-1). E=z−5(z10−1)z−1(z2−1)=z−4(z2−1)(z8+z6+z4+z2+1)z2−1=z−4(z8+z6+z4+z2+1)E = \frac{z^{-5}(z^{10}-1)}{z^{-1}(z^2-1)} = z^{-4}\frac{(z^2-1)(z^8+z^6+z^4+z^2+1)}{z^2-1} = z^{-4}(z^8+z^6+z^4+z^2+1). Distributing z−4z^{-4} gives: E=z4+z2+1+z−2+z−4E = z^4 + z^2 + 1 + z^{-2} + z^{-4}. Now, we group the terms: E=1+(z2+z−2)+(z4+z−4)E = 1 + (z^2 + z^{-2}) + (z^4 + z^{-4}). Using the identity zn+z−n=2cos⁡(nθ)z^n + z^{-n} = 2\cos(n\theta), we have: E=1+2cos⁡(2θ)+2cos⁡(4θ)E = 1 + 2\cos(2\theta) + 2\cos(4\theta). Distractor B is missing the constant term '1'. Distractor C is missing the factors of 2. Distractor D is an alternative, equivalent form of the expression using the identity zn−z−n=2isin⁡(nθ)z^n - z^{-n} = 2i\sin(n\theta), which leads to 2isin⁡(5θ)2isin⁡θ\frac{2i\sin(5\theta)}{2i\sin\theta}, but the question asks for a sum of cosines.