Complex Numbers & Trig - IB Mathematics Analysis and Approaches

Let $z_1$ and $z_2$ be the two roots of the equation $z^2 = 2+2i\sqrt{3}$ . Find the value of $|\text{arg}(z_1) - \text{arg}(z_2)|$ , where arguments are the principal arguments in $(-\pi, \pi]$ .
Answer: $\pi$ Geometrically, the two square roots of a non-zero complex number $w$ are of the form $z_1$ and $z_2 = -z_1$ . On the Argand diagram, they are separated by an angle of $\pi$ radians. Therefore, the difference in their arguments will be $\pi$ . To verify by calculation: First, convert $w = 2+2i\sqrt{3}$ to polar form. Modulus $|w| = \sqrt{2^2+(2\sqrt{3})^2} = \sqrt{4+12}=4$ . Argument $\text{Arg}(w) = \arctan(\frac{2\sqrt{3}}{2}) = \arctan(\sqrt{3}) = \frac{\pi}{3}$ . So $w = 4e^{i\pi/3}$ . The roots of $z^2 = w$ are given by $z_k = \sqrt{4}e^{i(\pi/3+2k\pi)/2}$ for $k=0,1$ . For $k=0$ : $z_1 = 2e^{i\pi/6}$ . The argument is $\text{arg}(z_1) = \frac{\pi}{6}$ . For $k=1$ : $z_2 = 2e^{i(\pi/6+\pi)} = 2e^{i7\pi/6}$ . The principal argument is $7\pi/6 - 2\pi = -5\pi/6$ . The difference is $|\frac{\pi}{6} - (-\frac{5\pi}{6})| = |\frac{6\pi}{6}| = \pi$ . Distractor B is the argument of the original number $w$ . Distractor A is the argument of one of the roots. Distractor C is double the argument of the original number.
Let $z = (\cos\alpha + i\sin\alpha)(\cos\alpha - i\sin\alpha)^{-3}$ . Which of the following is equal to $z$ ?
Answer: $\cos(4\alpha) + i\sin(4\alpha)$ We can write the expression in Euler's form. $\cos\alpha + i\sin\alpha = e^{i\alpha}$ and $\cos\alpha - i\sin\alpha = e^{-i\alpha}$ . The expression becomes: $z = (e^{i\alpha}) (e^{-i\alpha})^{-3}$ Using the laws of exponents: $z = e^{i\alpha} \cdot e^{(-i\alpha)(-3)} = e^{i\alpha} \cdot e^{i3\alpha} = e^{i\alpha + i3\alpha} = e^{i4\alpha}$ . Converting this back to trigonometric form gives: $z = \cos(4\alpha) + i\sin(4\alpha)$ . 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 $3-1=2$ instead of $1 - (-3) = 4$ .
The roots of the equation $(z - (1+i))^5 = 32$ are plotted on the Argand diagram. Which statement accurately describes their geometric arrangement?
Answer: The sum of the roots is $5+5i$ . Let $w = z - (1+i)$ . The equation becomes $w^5 = 32$ . The solutions for $w$ are the 5th roots of 32, given by $w_k = 2e^{i(2k\pi/5)}$ for $k=0,1,2,3,4$ . The sum of these roots of unity (scaled by 2) is $\sum_{k=0}^{4} w_k = 0$ . The roots for $z$ are given by $z_k = w_k + (1+i)$ . This means the set of roots for $z$ is a translation of the set of roots for $w$ by the vector corresponding to $1+i$ . The sum of the roots for $z$ is: $\sum_{k=0}^{4} z_k = \sum_{k=0}^{4} (w_k + (1+i)) = (\sum_{k=0}^{4} w_k) + \sum_{k=0}^{4} (1+i) = 0 + 5(1+i) = 5+5i$ . So, statement B is correct. Statement A is false: The roots form a regular pentagon centered at $1+i$ , not the origin. They lie on a circle of radius 2 centered at $1+i$ . Statement C is false: Expanding the equation leads to a polynomial whose constant term is $-(1+i)^5-32$ . The product of the roots is complex. Statement D is false: For a root $z_k$ to be real, its imaginary part must be zero. $z_k = (1+2\cos(2k\pi/5)) + i(1+2\sin(2k\pi/5))$ . For the imaginary part to be zero, $\sin(2k\pi/5) = -1/2$ . This does not occur for $k=0,1,2,3,4$ .
Find the root of $z^3 = 8i$ that lies in the second quadrant of the Argand diagram.
Answer: $-\sqrt{3} + i$ First, write $8i$ in polar form. The modulus is 8 and the argument is $\frac{\pi}{2}$ . So, $8i = 8e^{i(\pi/2 + 2k\pi)}$ . To find the cube roots, we take the cube root of the modulus and divide the argument by 3: $z = (8)^{1/3} e^{i(\pi/2 + 2k\pi)/3} = 2e^{i(\pi/6 + 2k\pi/3)}$ for $k=0, 1, 2$ . We find the three roots: For $k=0$ : $z_0 = 2e^{i\pi/6} = 2(\cos(\frac{\pi}{6}) + i\sin(\frac{\pi}{6})) = 2(\frac{\sqrt{3}}{2} + \frac{1}{2}i) = \sqrt{3} + i$ . This is in the first quadrant. For $k=1$ : $z_1 = 2e^{i(\pi/6 + 2\pi/3)} = 2e^{i5\pi/6} = 2(\cos(\frac{5\pi}{6}) + i\sin(\frac{5\pi}{6})) = 2(-\frac{\sqrt{3}}{2} + \frac{1}{2}i) = -\sqrt{3} + i$ . This is in the second quadrant. For $k=2$ : $z_2 = 2e^{i(\pi/6 + 4\pi/3)} = 2e^{i9\pi/6} = 2e^{i3\pi/2} = -2i$ . This lies on the negative imaginary axis. The root in the second quadrant is $-\sqrt{3} + i$ . Distractor A is a root of $z^3 = -8i$ . Distractor C is the root in the first quadrant. Distractor D is the root on the negative imaginary axis.
Given $z = \cos\theta + i\sin\theta$ , the expression $\frac{z^5 - z^{-5}}{z - z^{-1}}$ can be written as a sum of cosines. What is this sum?
Answer: $1 + 2\cos(2\theta) + 2\cos(4\theta)$ Let the expression be $E$ . We can recognize $z^5 - z^{-5}$ as $z^{-5}(z^{10}-1)$ and $z - z^{-1}$ as $z^{-1}(z^2-1)$ . $E = \frac{z^{-5}(z^{10}-1)}{z^{-1}(z^2-1)} = z^{-4}\frac{(z^2-1)(z^8+z^6+z^4+z^2+1)}{z^2-1} = z^{-4}(z^8+z^6+z^4+z^2+1)$ . Distributing $z^{-4}$ gives: $E = z^4 + z^2 + 1 + z^{-2} + z^{-4}$ . Now, we group the terms: $E = 1 + (z^2 + z^{-2}) + (z^4 + z^{-4})$ . Using the identity $z^n + z^{-n} = 2\cos(n\theta)$ , we have: $E = 1 + 2\cos(2\theta) + 2\cos(4\theta)$ . 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 $z^n - z^{-n} = 2i\sin(n\theta)$ , which leads to $\frac{2i\sin(5\theta)}{2i\sin\theta}$ , but the question asks for a sum of cosines.
Let $\omega$ be a non-real cube root of unity. What is the value of $(1 + \omega - \omega^2)^5$ ?
Answer: $-32\omega$ The cube roots of unity are the solutions to $z^3 - 1 = 0$ , which factors as $(z-1)(z^2+z+1)=0$ . Since $\omega$ is a non-real root, it must satisfy $\omega^2 + \omega + 1 = 0$ . From this identity, we can write $1 + \omega = -\omega^2$ . Substitute this into the expression: $(1 + \omega - \omega^2)^5 = (-\omega^2 - \omega^2)^5 = (-2\omega^2)^5$ . Now, we simplify this expression: $(-2\omega^2)^5 = (-2)^5 (\omega^2)^5 = -32 \omega^{10}$ . Since $\omega$ is a cube root of unity, $\omega^3 = 1$ . We can simplify $\omega^{10}$ as $\omega^{10} = (\omega^3)^3 \omega = 1^3 \omega = \omega$ . Therefore, the final value is $-32\omega$ . Distractor A results from incorrectly simplifying $\omega^{10}$ to 1. Distractor C may result from an error in simplifying $\omega^{10}$ or confusion with the other non-real root. Distractor D has a sign error in $(-2)^5$ .
The solutions to the equation $z^5 = -32$ form the vertices of a regular pentagon in the Argand diagram. What is the area of this pentagon?
Answer: $10\sin(\frac{2\pi}{5})$ The equation is $z^5 = -32$ . In polar form, $-32 = 32e^{i\pi}$ . The solutions are $z_k = (32e^{i(\pi+2k\pi)})^{1/5} = 2e^{i(\pi+2k\pi)/5}$ for $k=0,1,2,3,4$ . All five roots have a modulus of $r=2$ , 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 $r=2$ and the angle between them being $\frac{2\pi}{5}$ . The area of one such triangle is $\frac{1}{2}ab\sin C = \frac{1}{2}(2)(2)\sin(\frac{2\pi}{5}) = 2\sin(\frac{2\pi}{5})$ . The total area of the pentagon is 5 times the area of one triangle: Area $= 5 \times 2\sin(\frac{2\pi}{5}) = 10\sin(\frac{2\pi}{5})$ . Distractor A uses an incorrect central angle of $\pi/5$ . Distractor D forgets the $1/2$ in the triangle area formula. Distractor C makes both of these errors.
Using De Moivre's theorem, an expression for $\cos(4\theta)$ can be found by considering the real part of $(\cos\theta + i\sin\theta)^4$ . Which of the following is the correct expression in terms of $\cos\theta$ ?
Answer: $8\cos^4\theta - 8\cos^2\theta + 1$ By De Moivre's theorem, $\cos(4\theta) + i\sin(4\theta) = (\cos\theta + i\sin\theta)^4$ . We expand the right side using the binomial theorem: $(\cos\theta + i\sin\theta)^4 = \binom{4}{0}\cos^4\theta + \binom{4}{1}\cos^3\theta(i\sin\theta) + \binom{4}{2}\cos^2\theta(i\sin\theta)^2 + \binom{4}{3}\cos\theta(i\sin\theta)^3 + \binom{4}{4}(i\sin\theta)^4$ $= \cos^4\theta + 4i\cos^3\theta\sin\theta - 6\cos^2\theta\sin^2\theta - 4i\cos\theta\sin^3\theta + \sin^4\theta$ The real part is $\cos(4\theta) = \cos^4\theta - 6\cos^2\theta\sin^2\theta + \sin^4\theta$ . To express this in terms of $\cos\theta$ , we substitute $\sin^2\theta = 1 - \cos^2\theta$ : $\cos(4\theta) = \cos^4\theta - 6\cos^2\theta(1 - \cos^2\theta) + (1 - \cos^2\theta)^2$ $= \cos^4\theta - 6\cos^2\theta + 6\cos^4\theta + (1 - 2\cos^2\theta + \cos^4\theta)$ $= (1+6+1)\cos^4\theta + (-6-2)\cos^2\theta + 1$ $= 8\cos^4\theta - 8\cos^2\theta + 1$ . Distractor A is the expression for $\cos(3\theta)$ . Distractor B has an algebraic error in collecting the $\cos^2\theta$ terms. Distractor D is the correct real part before converting sine terms to cosine terms.
Let $z = \cos\theta + i\sin\theta$ . The expression for $-4\sin^3\theta$ can be derived from the binomial expansion of $(z-z^{-1})^3$ . What is this expression in terms of sines of multiple angles?
Answer: $\sin(3\theta) - 3\sin\theta$ We use the identity $z - z^{-1} = 2i\sin\theta$ . Cubing both sides gives $(z-z^{-1})^3 = (2i\sin\theta)^3 = 8i^3\sin^3\theta = -8i\sin^3\theta$ . We also expand $(z-z^{-1})^3$ binomially: $(z-z^{-1})^3 = z^3 - 3z^2(z^{-1}) + 3z(z^{-1})^2 - (z^{-1})^3 = z^3 - 3z + 3z^{-1} - z^{-3}$ . Grouping terms: $(z^3-z^{-3}) - 3(z-z^{-1})$ . Using the identity $z^n-z^{-n} = 2i\sin(n\theta)$ , this becomes: $2i\sin(3\theta) - 3(2i\sin\theta) = 2i(\sin(3\theta) - 3\sin\theta)$ . Equating the two expressions for $(z-z^{-1})^3$ : $-8i\sin^3\theta = 2i(\sin(3\theta) - 3\sin\theta)$ . Dividing both sides by $-2i$ gives: $4\sin^3\theta = -(\sin(3\theta) - 3\sin\theta) = 3\sin\theta - \sin(3\theta)$ . The question asks for $-4\sin^3\theta$ , so we multiply by -1: $-4\sin^3\theta = -(3\sin\theta - \sin(3\theta)) = \sin(3\theta) - 3\sin\theta$ . Distractor A is the expression for $+4\sin^3\theta$ . Distractors C and D are incorrect expressions involving cosines.
Find the value of the sum $S = \sum_{k=0}^{10} \cos\left(\frac{k\pi}{5}\right)$ .
Answer: 1 We consider the sum of a complex geometric series $Z = \sum_{k=0}^{10} e^{ik\pi/5}$ . The real part of this sum will be the answer, $S = \text{Re}(Z)$ . The series is $Z = \sum_{k=0}^{10} (e^{i\pi/5})^k$ . This is a geometric series with first term $a=1$ , common ratio $r = e^{i\pi/5}$ , and $n=11$ terms. The sum is given by $S_n = a\frac{r^n-1}{r-1}$ . $Z = 1 \cdot \frac{(e^{i\pi/5})^{11} - 1}{e^{i\pi/5} - 1} = \frac{e^{i11\pi/5} - 1}{e^{i\pi/5} - 1}$ . We simplify the term in the numerator: $e^{i11\pi/5} = e^{i(10\pi/5 + \pi/5)} = e^{i(2\pi + \pi/5)} = e^{i2\pi}e^{i\pi/5} = 1 \cdot e^{i\pi/5} = e^{i\pi/5}$ . Substituting this back into the sum: $Z = \frac{e^{i\pi/5} - 1}{e^{i\pi/5} - 1} = 1$ . The sum of the complex numbers is $1$ , which can be written as $1+0i$ . The real part is $\text{Re}(Z) = 1$ . Distractor B is a common result for sums of roots of unity, but the limits of this summation are different. Distractor A is another plausible but incorrect result for such sums.
Simplify the expression $(\cos\theta - i\sin\theta)^5(\cos\theta + i\sin\theta)^{-3}$ .
Answer: $\cos(8\theta) - i\sin(8\theta)$ We can write the expression using Euler's form. $\cos\theta + i\sin\theta = e^{i\theta}$ $\cos\theta - i\sin\theta = \cos(-\theta) + i\sin(-\theta) = e^{-i\theta}$ The expression becomes: $(e^{-i\theta})^5 (e^{i\theta})^{-3}$ Using the laws of exponents, this simplifies to: $e^{-i5\theta} \cdot e^{-i3\theta} = e^{-i5\theta - i3\theta} = e^{-i8\theta}$ Converting back to polar form: $e^{-i8\theta} = \cos(-8\theta) + i\sin(-8\theta)$ Since cosine is an even function ( $\cos(-x)=\cos(x)$ ) and sine is an odd function ( $\sin(-x)=-\sin(x)$ ), this is: $\cos(8\theta) - i\sin(8\theta)$ . Distractor A results from incorrectly combining exponents as $5-3=2$ and ignoring the initial negative sign. Distractor B also incorrectly combines exponents as $5-3=2$ . Distractor C has the correct angle but the wrong sign, arising from an error with the negative exponents.
Let $z = \frac{1}{2} - \frac{\sqrt{3}}{2}i$ . Find the value of $z^{2023}$ .
Answer: $\frac{1}{2} - \frac{\sqrt{3}}{2}i$ First, convert $z$ to its polar form, $r(\cos\theta + i\sin\theta)$ or $re^{i\theta}$ . The modulus is $r = |z| = \sqrt{(\frac{1}{2})^2 + (-\frac{\sqrt{3}}{2})^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = 1$ . The argument is $\theta = \arctan\left(\frac{-\sqrt{3}/2}{1/2}\right) = \arctan(-\sqrt{3}) = -\frac{\pi}{3}$ (since $z$ is in the fourth quadrant). So, $z = 1 \cdot e^{-i\pi/3} = e^{-i\pi/3}$ . Using De Moivre's theorem, $z^{2023} = (e^{-i\pi/3})^{2023} = e^{-i2023\pi/3}$ . To simplify the argument, we find the remainder of 2023 divided by 6 (the period of $e^{ik\pi/3}$ ). Or we can write $-2023\pi/3$ as $(-674 \cdot 3 - 1)\pi/3 = -674\pi - \pi/3$ . Since $-674\pi$ is an integer multiple of $2\pi$ , the angle is coterminal with $-\pi/3$ . So, $z^{2023} = e^{-i\pi/3} = z = \frac{1}{2} - \frac{\sqrt{3}}{2}i$ . 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 $z$ .
Let $\omega$ be a non-real 7th root of unity. What is the value of the sum $S = \sum_{k=0}^{6} (1+\omega^k)^2$ ?
Answer: 7 We expand the term inside the summation first: $S = \sum_{k=0}^{6} (1 + 2\omega^k + \omega^{2k})$ . We can split the summation into three parts: $S = \sum_{k=0}^{6} 1 + \sum_{k=0}^{6} 2\omega^k + \sum_{k=0}^{6} \omega^{2k}$ $S = \sum_{k=0}^{6} 1 + 2\sum_{k=0}^{6} \omega^k + \sum_{k=0}^{6} (\omega^2)^k$ . For the $n$ -th roots of unity, we know that $\sum_{k=0}^{n-1} \omega^k = 0$ . Here, $n=7$ , so $\sum_{k=0}^{6} \omega^k = 0$ . The first term is $\sum_{k=0}^{6} 1 = 7$ . The second term is $2\sum_{k=0}^{6} \omega^k = 2(0) = 0$ . For the third term, let $u = \omega^2$ . Since $\omega$ is a non-real 7th root of unity, $\omega^7=1$ and $\omega \neq 1$ . Then $u^7 = (\omega^2)^7 = (\omega^7)^2 = 1^2 = 1$ . Also, $u \neq 1$ since $\omega^2$ is also a non-real 7th root of unity. So the sum $\sum_{k=0}^{6} u^k = \sum_{k=0}^{6} (\omega^2)^k = 0$ . Therefore, $S = 7 + 0 + 0 = 7$ . Distractor A incorrectly assumes the entire sum is zero. Distractor C comes from perhaps thinking the third sum is 7, so $7+0+7=14$ . Distractor D might come from $7+7+7=21$ .
Given that $z^n + z^{-n} = 2\cos(n\theta)$ , which of the following is equivalent to $16\cos^5\theta$ ?
Answer: $10\cos\theta + 5\cos(3\theta) + \cos(5\theta)$ We start with $2\cos\theta = z + z^{-1}$ and raise both sides to the power of 5. $(2\cos\theta)^5 = 32\cos^5\theta = (z + z^{-1})^5$ . Using the binomial expansion for $(z+z^{-1})^5$ , with coefficients 1, 5, 10, 10, 5, 1: $z^5 + 5z^4(z^{-1}) + 10z^3(z^{-1})^2 + 10z^2(z^{-1})^3 + 5z(z^{-1})^4 + z^{-5}$ $= z^5 + 5z^3 + 10z + 10z^{-1} + 5z^{-3} + z^{-5}$ Group the terms: $= (z^5 + z^{-5}) + 5(z^3 + z^{-3}) + 10(z + z^{-1})$ Substitute the identity $z^n + z^{-n} = 2\cos(n\theta)$ : $32\cos^5\theta = 2\cos(5\theta) + 5(2\cos(3\theta)) + 10(2\cos\theta)$ $32\cos^5\theta = 2\cos(5\theta) + 10\cos(3\theta) + 20\cos\theta$ To find $16\cos^5\theta$ , we divide by 2: $16\cos^5\theta = \cos(5\theta) + 5\cos(3\theta) + 10\cos\theta$ . Distractor D has the coefficients mixed up. Distractor C is the expression for $32\cos^5\theta$ but with an error. Distractor B uses sine functions instead of cosine functions.
Let $z_k$ for $k=0, 1, 2, 3, 4$ be the five roots of the equation $z^5 = 1-i$ . What is the value of the product of these roots, $P = z_0 z_1 z_2 z_3 z_4$ ?
Answer: $1-i$ The equation is $z^5 = 1-i$ , which can be written as a polynomial equation $z^5 - (1-i) = 0$ . For a polynomial of the form $a_n z^n + a_{n-1} z^{n-1} + ... + a_0 = 0$ , the product of the roots is given by $(-1)^n \frac{a_0}{a_n}$ . In this case, $n=5$ , the leading coefficient is $a_5 = 1$ , and the constant term is $a_0 = -(1-i)$ . The product of the roots is $P = (-1)^5 \frac{-(1-i)}{1} = (-1)(-(1-i)) = 1-i$ . Distractor A results from a sign error, forgetting the $(-1)^5$ factor. Distractor C is the conjugate of the correct answer, a common error. Distractor D is the modulus of the product, $|1-i| = \sqrt{1^2+(-1)^2} = \sqrt{2}$ , not the product itself.
If $z = \cos\theta + i\sin\theta$ , which of the following real expressions is equivalent to $\frac{(z^2+1)^2}{z^2}$ ?
Answer: $4\cos^2\theta$ We can rewrite the expression by dividing the numerator by the denominator: $\frac{(z^2+1)^2}{z^2} = \frac{z^4+2z^2+1}{z^2} = z^2 + 2 + z^{-2}$ . Now, we group the terms: $(z^2 + z^{-2}) + 2$ . Using the identity $z^n + z^{-n} = 2\cos(n\theta)$ , we have: $z^2 + z^{-2} = 2\cos(2\theta)$ . So the expression becomes $2\cos(2\theta) + 2$ . Using the double angle identity for cosine, $\cos(2\theta) = 2\cos^2\theta - 1$ , we can simplify further: $2(2\cos^2\theta - 1) + 2 = 4\cos^2\theta - 2 + 2 = 4\cos^2\theta$ . 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.
Consider the equation $(z+1)^4 = z^4$ . Which of the following statements is true about its roots in the complex plane?
Answer: All roots lie on the line $\text{Re}(z) = -\frac{1}{2}$ . The equation can be rewritten as $\left(\frac{z+1}{z}\right)^4 = 1$ . Let $w = \frac{z+1}{z}$ . The equation becomes $w^4 = 1$ . The solutions for $w$ are the 4th roots of unity: $1, i, -1, -i$ . We have $w = 1 + \frac{1}{z}$ . The case $w=1$ implies $1 = 1 + \frac{1}{z}$ , which means $\frac{1}{z} = 0$ , an impossibility. So there are only three roots for $z$ , corresponding to $w = i, -1, -i$ . We solve for $z$ using $z = \frac{1}{w-1}$ : If $w = i$ : $z = \frac{1}{i-1} = \frac{-1-i}{2} = -\frac{1}{2} - \frac{1}{2}i$ . If $w = -1$ : $z = \frac{1}{-1-1} = -\frac{1}{2}$ . If $w = -i$ : $z = \frac{1}{-i-1} = \frac{-1+i}{2} = -\frac{1}{2} + \frac{1}{2}i$ . All three roots have a real part of $-\frac{1}{2}$ , so they lie on the vertical line $\text{Re}(z) = -\frac{1}{2}$ . Alternatively, expanding $(z+1)^4=z^4$ gives $4z^3+6z^2+4z+1=0$ . The sum of roots is $-\frac{6}{4} = -\frac{3}{2}$ (A is false). The product of roots is $-\frac{1}{4}$ (B is false). D is false as two roots are complex.
Let $z = 1 + \cos(2\theta) + i\sin(2\theta)$ for $0 < \theta < \pi/2$ . What is the modulus of $z$ ?
Answer: $2\cos\theta$ We can simplify the expression for $z$ using double angle identities. $1 + \cos(2\theta) = 2\cos^2\theta$ $\sin(2\theta) = 2\sin\theta\cos\theta$ Substituting these into the expression for $z$ : $z = 2\cos^2\theta + i(2\sin\theta\cos\theta)$ We can factor out $2\cos\theta$ : $z = 2\cos\theta (\cos\theta + i\sin\theta)$ This is now in the form $r(\cos\alpha + i\sin\alpha)$ , which is the polar form. The modulus $r$ is the term multiplying the cis part. Since $0 < \theta < \pi/2$ , $\cos\theta > 0$ , so $2\cos\theta$ is positive. Therefore, the modulus is $|z| = 2\cos\theta$ . Distractor A would arise from using an incorrect identity for $1+\cos(2\theta)$ . Distractors C and D are incorrect applications of trigonometric identities.
Let $z = \cos\theta + i\sin\theta$ . Using the identity $2\cos\theta = z + z^{-1}$ , find the expression for $8\cos^3\theta$ in terms of multiple angles.
Answer: $2\cos(3\theta) + 6\cos\theta$ We start with $2\cos\theta = z + z^{-1}$ . Cubing both sides gives: $(2\cos\theta)^3 = (z + z^{-1})^3$ $8\cos^3\theta = z^3 + 3z^2(z^{-1}) + 3z(z^{-1})^2 + (z^{-1})^3$ $8\cos^3\theta = z^3 + 3z + 3z^{-1} + z^{-3}$ Group the terms: $8\cos^3\theta = (z^3 + z^{-3}) + 3(z + z^{-1})$ . Using the general identity $z^n + z^{-n} = 2\cos(n\theta)$ , we substitute: $8\cos^3\theta = (2\cos(3\theta)) + 3(2\cos\theta)$ $8\cos^3\theta = 2\cos(3\theta) + 6\cos\theta$ . Distractor A arises from forgetting the factors of 2 in the identities $z^n+z^{-n}=2\cos(n\theta)$ . Distractor C would result from expanding $(z-z^{-1})^3$ but incorrectly matching it to cosine terms. Distractor D results from incorrectly using sine identities.
If $z = e^{i\theta}$ , what is the modulus of $1+z+z^2$ ?
Answer: $\left| \frac{\sin(3\theta/2)}{\sin(\theta/2)} \right|$ We can express $1+z+z^2$ as the sum of a geometric series: $1+z+z^2 = \frac{z^3-1}{z-1}$ . Therefore, $|1+z+z^2| = \frac{|z^3-1|}{|z-1|}$ . Let's evaluate $|z^n-1|$ where $z=e^{i\theta}$ . $|z^n-1|^2 = (z^n-1)(\overline{z^n}-1) = (z^n-1)(z^{-n}-1) = z^n z^{-n} - z^n - z^{-n} + 1 = 1 - (z^n+z^{-n}) + 1 = 2 - 2\cos(n\theta)$ . Using the half-angle identity $1-\cos(A) = 2\sin^2(A/2)$ , this becomes $2(1-\cos(n\theta)) = 4\sin^2(n\theta/2)$ . So, $|z^n-1| = 2|\sin(n\theta/2)|$ . Applying this to our expression: $|1+z+z^2| = \frac{2|\sin(3\theta/2)|}{2|\sin(\theta/2)|} = \left| \frac{\sin(3\theta/2)}{\sin(\theta/2)} \right|$ . Distractor A is the result of incorrectly applying the triangle inequality: $|1+z+z^2| \leq |1|+|z|+|z^2| = 1+1+1=3$ . Distractor B uses the wrong half-angle identity (for $1+\cos A$ ). Distractor C makes an error in the half-angle formula application.

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