Polar & Exponential Form - IB Mathematics Analysis and Approaches

Let $z = 1 - \cos(2\alpha) - i\sin(2\alpha)$ , for $0 < \alpha < \pi/2$ . Find the modulus of $z$ .
Answer: $2\sin\alpha$ We can find the modulus using the formula $|z| = \sqrt{\operatorname{Re}(z)^2 + \operatorname{Im}(z)^2}$ . Here, $\operatorname{Re}(z) = 1 - \cos(2\alpha)$ and $\operatorname{Im}(z) = -\sin(2\alpha)$ . So, $|z|^2 = (1-\cos(2\alpha))^2 + (-\sin(2\alpha))^2 = 1 - 2\cos(2\alpha) + \cos^2(2\alpha) + \sin^2(2\alpha)$ . Since $\cos^2(2\alpha) + \sin^2(2\alpha) = 1$ , this simplifies to $|z|^2 = 1 - 2\cos(2\alpha) + 1 = 2 - 2\cos(2\alpha) = 2(1-\cos(2\alpha))$ . Using the half-angle identity $1-\cos(2\alpha) = 2\sin^2\alpha$ , we get $|z|^2 = 2(2\sin^2\alpha) = 4\sin^2\alpha$ . Therefore, $|z| = \sqrt{4\sin^2\alpha} = 2|\sin\alpha|$ . Since $0 < \alpha < \pi/2$ , $\sin\alpha$ is positive, so $|z| = 2\sin\alpha$ .
Let $z = \cos\theta + i\sin\theta$ . For what value of $\theta$ in the range $0 \le \theta < 2\pi$ is the real part of $\frac{1}{1-z}$ equal to $\frac{1}{2}$ ?
Answer: Any $\theta \neq 0$ Let $z = \cos\theta + i\sin\theta$ . We analyze the expression $\frac{1}{1-z}$ . Substitute for z: $\frac{1}{1 - (\cos\theta + i\sin\theta)} = \frac{1}{(1-\cos\theta) - i\sin\theta}$ . To find the real part, multiply the numerator and denominator by the conjugate of the denominator: $\frac{(1-\cos\theta) + i\sin\theta}{((1-\cos\theta) - i\sin\theta)((1-\cos\theta) + i\sin\theta)} = \frac{1-\cos\theta + i\sin\theta}{(1-\cos\theta)^2 + \sin^2\theta}$ . Let's simplify the denominator: $1 - 2\cos\theta + \cos^2\theta + \sin^2\theta = 1 - 2\cos\theta + 1 = 2 - 2\cos\theta = 2(1-\cos\theta)$ . The expression becomes $\frac{1-\cos\theta + i\sin\theta}{2(1-\cos\theta)}$ . The real part is $\operatorname{Re}\left(\frac{1}{1-z}\right) = \frac{1-\cos\theta}{2(1-\cos\theta)}$ . As long as $1-\cos\theta \neq 0$ (i.e., $\cos\theta \neq 1$ , which means $\theta \neq 0$ in the given range), we can cancel the term $1-\cos\theta$ , and the real part is $\frac{1}{2}$ . Therefore, the condition is met for any $\theta$ in the range $0 \le \theta < 2\pi$ except for $\theta=0$ .
Given two non-zero complex numbers $z_1$ and $z_2$ with principal arguments $\arg(z_1) = \alpha$ and $\arg(z_2) = \beta$ , where $\frac{\pi}{2} < \alpha < \pi$ and $-\frac{\pi}{2} < \beta < 0$ . Which of the following statements must be true about their product $z_1 z_2$ ?
Answer: $\operatorname{Im}(z_1 z_2) > 0$ The argument of the product is the sum of the arguments: $\arg(z_1 z_2) = \arg(z_1) + \arg(z_2) = \alpha + \beta$ . We need to find the range of $\alpha + \beta$ . Given $\frac{\pi}{2} < \alpha < \pi$ and $-\frac{\pi}{2} < \beta < 0$ . The minimum value of the sum is $\frac{\pi}{2} + (-\frac{\pi}{2}) = 0$ . The maximum value of the sum is $\pi + 0 = \pi$ . Therefore, $0 < \alpha + \beta < \pi$ . Let $\theta = \arg(z_1 z_2)$ . We have $0 < \theta < \pi$ . For any complex number $w = |w|(\%cos\theta + i\sin\theta)$ , its imaginary part is $|w|\sin\theta$ . Since $z_1$ and $z_2$ are non-zero, $|w| > 0$ . In the interval $0 < \theta < \pi$ , $\sin\theta > 0$ . Thus, $\operatorname{Im}(z_1 z_2) > 0$ . The real part, $|w|\cos\theta$ , can be positive (if $0 < \theta < \pi/2$ ) or negative (if $\pi/2 < \theta < \pi$ ), so we cannot determine the sign of the real part.
The three distinct cube roots of $z = 8i$ form the vertices of a triangle in the Argand plane. What is the area of this triangle?
Answer: $3\sqrt{3}$ First, express $z=8i$ in polar form. The modulus is $r=|8i|=8$ . The argument is $\theta = \frac{\pi}{2}$ . So $z = 8e^{i\pi/2}$ . The cube roots $w_k$ are given by $w_k = \sqrt[3]{8} e^{i(\frac{\pi/2 + 2k\pi}{3})}$ for $k=0, 1, 2$ . The modulus of each root is $R = \sqrt[3]{8} = 2$ . 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 $A = \frac{1}{2}nR^2\sin(\frac{2\pi}{n})$ . For an equilateral triangle, $n=3$ , so the area is $A = \frac{1}{2}(3)(2^2)\sin(\frac{2\pi}{3}) = \frac{3}{2}(4)(\frac{\sqrt{3}}{2}) = 3\sqrt{3}$ .
Which of the following is a solution to the equation $e^z = \sqrt{3} - i$ ?
Answer: $\ln 2 - i\frac{\pi}{6}$ Let $z = x+iy$ . The equation is $e^{x+iy} = \sqrt{3} - i$ . This can be written as $e^x e^{iy} = \sqrt{3} - i$ . We need to write the right side in exponential form, $re^{i\theta}$ . The modulus is $r = |\sqrt{3} - i| = \sqrt{(\sqrt{3})^2 + (-1)^2} = \sqrt{3+1} = 2$ . The argument $\theta$ is in the fourth quadrant, so $\theta = \arctan(\frac{-1}{\sqrt{3}}) = -\frac{\pi}{6}$ . Thus, $\sqrt{3} - i = 2e^{-i\pi/6}$ . The equation becomes $e^x e^{iy} = 2e^{i(-\pi/6 + 2k\pi)}$ for any integer $k$ . By equating the moduli, we get $e^x = 2$ , which implies $x = \ln 2$ . By equating the arguments, we get $y = -\frac{\pi}{6} + 2k\pi$ . For $k=0$ , we get a solution $z = x+iy = \ln 2 - i\frac{\pi}{6}$ .
Let $z = e^{i\theta}$ . The expression $\frac{z^2 - 1}{z^2 + 1}$ simplifies to which of the following?
Answer: $i\tan\theta$ Let $z=e^{i\theta}$ . Then $z^2 = e^{i2\theta}$ . Substitute this into the expression: $\frac{e^{i2\theta} - 1}{e^{i2\theta} + 1}$ . To simplify, multiply the numerator and denominator by $e^{-i\theta}$ : $\frac{e^{-i\theta}(e^{i2\theta} - 1)}{e^{-i\theta}(e^{i2\theta} + 1)} = \frac{e^{i\theta} - e^{-i\theta}}{e^{i\theta} + e^{-i\theta}}$ . Using Euler's relations, $\cos\theta = \frac{e^{i\theta}+e^{-i\theta}}{2}$ and $\sin\theta = \frac{e^{i\theta}-e^{-i\theta}}{2i}$ , we can rewrite the numerator as $2i\sin\theta$ and the denominator as $2\cos\theta$ . The expression becomes $\frac{2i\sin\theta}{2\cos\theta} = i\tan\theta$ .
The locus of a point $z$ in the complex plane is given by $|z - 4| = |z - 2i|$ . Find the argument of the point on this locus that is closest to the origin.
Answer: $\arctan(-1/2)$ The equation $|z-z_1| = |z-z_2|$ describes the perpendicular bisector of the line segment connecting $z_1$ and $z_2$ . Here, $z_1 = 4$ and $z_2 = 2i$ . The segment connects the points (4, 0) and (0, 2). The midpoint of this segment is $\left(\frac{4+0}{2}, \frac{0+2}{2}\right) = (2, 1)$ . The slope of the segment is $\frac{2-0}{0-4} = -\frac{1}{2}$ . 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: $y-1 = 2(x-2) \Rightarrow y = 2x - 3$ . 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 $-\frac{1}{2}$ , so its equation is $y = -\frac{1}{2}x$ . To find the intersection point, we set the y-values equal: $2x - 3 = -\frac{1}{2}x \Rightarrow 4x - 6 = -x \Rightarrow 5x = 6 \Rightarrow x = \frac{6}{5}$ . Then $y = -\frac{1}{2}(\frac{6}{5}) = -\frac{3}{5}$ . The point is $z = \frac{6}{5} - \frac{3}{5}i$ . The argument is $\arg(z) = \arctan\left(\frac{-3/5}{6/5}\right) = \arctan(-\frac{1}{2})$ .
A complex number $z$ has modulus $r$ and argument $\theta$ . The complex number $w = \frac{z}{\bar{z}} - \frac{\bar{z}}{z}$ is purely imaginary. What is its imaginary part?
Answer: $2\sin(2\theta)$ Let $z = re^{i\theta}$ . Then its conjugate is $\bar{z} = re^{-i\theta}$ . Now we can evaluate the terms in the expression for $w$ . $\frac{z}{\bar{z}} = \frac{re^{i\theta}}{re^{-i\theta}} = e^{i\theta - (-i\theta)} = e^{i2\theta}$ . $\frac{\bar{z}}{z} = \frac{re^{-i\theta}}{re^{i\theta}} = e^{-i\theta - i\theta} = e^{-i2\theta}$ . So, $w = e^{i2\theta} - e^{-i2\theta}$ . Using Euler's formula, $e^{i\phi} = \cos\phi + i\sin\phi$ , we have $w = (\cos(2\theta) + i\sin(2\theta)) - (\cos(-2\theta) + i\sin(-2\theta))$ . This simplifies to $w = (\cos(2\theta) + i\sin(2\theta)) - (\cos(2\theta) - i\sin(2\theta)) = 2i\sin(2\theta)$ . This is a purely imaginary number. The imaginary part is the coefficient of $i$ , which is $2\sin(2\theta)$ . Note that $2i\sin(2\theta)$ is the complex number itself, not its imaginary part.
Let $z = \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2}$ . If a complex number $w$ is multiplied by $z^3$ , what is the geometric transformation applied to $w$ in the Argand diagram?
Answer: Rotation by $135^\circ$ counter-clockwise about the origin First, we express $z$ in polar form. The modulus is $|z| = \sqrt{(\frac{\sqrt{2}}{2})^2 + (\frac{\sqrt{2}}{2})^2} = \sqrt{\frac{2}{4} + \frac{2}{4}} = \sqrt{1} = 1$ . The argument is $\arg(z) = \arctan\left(\frac{\sqrt{2}/2}{\sqrt{2}/2}\right) = \arctan(1) = \frac{\pi}{4}$ or $45^\circ$ . So, $z = e^{i\pi/4}$ . Next, we find $z^3$ using De Moivre's theorem: $z^3 = (e^{i\pi/4})^3 = e^{i3\pi/4}$ . The modulus of $z^3$ is 1 and its argument is $\frac{3\pi}{4}$ , which is $135^\circ$ . Multiplying a complex number $w$ by a complex number $\rho e^{i\theta}$ corresponds to scaling $w$ by a factor of $\rho$ and rotating it counter-clockwise by an angle of $\theta$ about the origin. Here, $z^3$ has modulus $\rho=1$ and argument $\theta = 135^\circ$ . Thus, the transformation is a rotation by $135^\circ$ counter-clockwise about the origin with no scaling.
What is the argument of the complex number $z = 1 - \cos(2\alpha) - i\sin(2\alpha)$ for $0 < \alpha < \pi/2$ ?
Answer: $\alpha - \frac{\pi}{2}$ We can factor the expression for $z$ . Using half-angle identities, $1-\cos(2\alpha) = 2\sin^2\alpha$ and $\sin(2\alpha) = 2\sin\alpha\cos\alpha$ . Substitute these into $z$ : $z = 2\sin^2\alpha - i(2\sin\alpha\cos\alpha)$ . Since $0 < \alpha < \pi/2$ , $\sin\alpha > 0$ , so we can factor out $2\sin\alpha$ : $z = 2\sin\alpha(\sin\alpha - i\cos\alpha)$ . The modulus is $r = 2\sin\alpha$ . The argument is determined by the term in the parenthesis. Let's convert this to polar form. We want to write $\sin\alpha - i\cos\alpha$ as $\cos\theta + i\sin\theta$ . We can use the identities $\sin\alpha = \cos(\frac{\pi}{2}-\alpha)$ and $\cos\alpha = \sin(\frac{\pi}{2}-\alpha)$ . So, $\sin\alpha - i\cos\alpha = \cos(\frac{\pi}{2}-\alpha) - i\sin(\frac{\pi}{2}-\alpha)$ . This is equal to $\cos(-(\frac{\pi}{2}-\alpha)) + i\sin(-(\frac{\pi}{2}-\alpha)) = \cos(\alpha-\frac{\pi}{2}) + i\sin(\alpha-\frac{\pi}{2})$ . Thus, the argument of $z$ is $\alpha - \frac{\pi}{2}$ .
Consider the complex number $w = \frac{1}{2}i$ . Find the value of the infinite sum $S = \sum_{n=0}^{\infty} w^n$ .
Answer: $\frac{4}{5} + \frac{2}{5}i$ This is an infinite geometric series with first term $a=w^0=1$ and common ratio $r=w=\frac{1}{2}i$ . The sum converges if $|r|<1$ . Here, $|r| = |\frac{1}{2}i| = \frac{1}{2} < 1$ , so the sum exists and is given by the formula $S = \frac{a}{1-r}$ . Substituting the values, we get $S = \frac{1}{1 - \frac{1}{2}i}$ . To express this in Cartesian form $x+iy$ , we multiply the numerator and denominator by the conjugate of the denominator: $S = \frac{1}{1 - \frac{1}{2}i} \times \frac{1 + \frac{1}{2}i}{1 + \frac{1}{2}i} = \frac{1 + \frac{1}{2}i}{1^2 - (\frac{1}{2}i)^2} = \frac{1 + \frac{1}{2}i}{1 - (-\frac{1}{4})} = \frac{1 + \frac{1}{2}i}{1 + \frac{1}{4}} = \frac{1 + \frac{1}{2}i}{\frac{5}{4}}$ . This simplifies to $S = \frac{4}{5}(1 + \frac{1}{2}i) = \frac{4}{5} + \frac{2}{5}i$ .
Let $\omega$ be a non-real cube root of unity. What is the value of the expression $(1 - \omega + \omega^2)(1 + \omega - \omega^2)$ ?
Answer: 4 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 be a root of $z^2+z+1=0$ . This gives us the crucial identity $1+\omega+\omega^2=0$ . From this, we can derive $1+\omega^2 = -\omega$ and $1+\omega = -\omega^2$ . Now we simplify the given expression. The first factor is $1 - \omega + \omega^2 = (1+\omega^2) - \omega = -\omega - \omega = -2\omega$ . The second factor is $1 + \omega - \omega^2 = (1+\omega) - \omega^2 = -\omega^2 - \omega^2 = -2\omega^2$ . The product is $(-2\omega)(-2\omega^2) = 4\omega^3$ . Since $\omega$ is a cube root of unity, $\omega^3 = 1$ . Therefore, the value of the expression is $4(1) = 4$ .
Given that $z_1 = r_1 e^{i\theta_1}$ and $z_2 = r_2 e^{i\theta_2}$ are such that $z_1 z_2$ is purely real and $\frac{z_1}{z_2}$ is purely imaginary, which of the following is a possible value for $\theta_1$ ?
Answer: $-\frac{\pi}{4}$ The product is $z_1 z_2 = r_1 r_2 e^{i(\theta_1+\theta_2)}$ . For this to be purely real, its argument must be a multiple of $\pi$ . So, $\theta_1+\theta_2 = k\pi$ for some integer $k$ . The quotient is $\frac{z_1}{z_2} = \frac{r_1}{r_2} e^{i(\theta_1-\theta_2)}$ . For this to be purely imaginary, its argument must be an odd multiple of $\frac{\pi}{2}$ . So, $\theta_1-\theta_2 = \frac{(2m+1)\pi}{2}$ for some integer $m$ . We have a system of two equations: (1) $\theta_1+\theta_2 = k\pi$ and (2) $\theta_1-\theta_2 = (m+\frac{1}{2})\pi$ . Adding the two equations gives $2\theta_1 = k\pi + (m+\frac{1}{2})\pi = (k+m+\frac{1}{2})\pi$ . Therefore, $\theta_1 = \frac{(k+m+\frac{1}{2})\pi}{2} = \frac{(2(k+m)+1)\pi}{4}$ . This means $\theta_1$ must be an odd multiple of $\frac{\pi}{4}$ . Among the given options, only $-\frac{\pi}{4}$ fits this condition (for $2(k+m)+1 = -1$ ).
If $z = \cos\theta + i\sin\theta$ , which of the following expressions is equivalent to $z^n + z^{-n}$ ?
Answer: $2\cos(n\theta)$ Given $z = \cos\theta + i\sin\theta$ , we can use De Moivre's theorem to find $z^n$ and $z^{-n}$ . According to the theorem, $z^n = (\cos\theta + i\sin\theta)^n = \cos(n\theta) + i\sin(n\theta)$ . Similarly, $z^{-n} = (\cos\theta + i\sin\theta)^{-n} = \cos(-n\theta) + i\sin(-n\theta)$ . Using the properties of trigonometric functions, $\cos(-x) = \cos(x)$ and $\sin(-x) = -\sin(x)$ , we can simplify $z^{-n}$ to $\cos(n\theta) - i\sin(n\theta)$ . Now, we add the two expressions: $z^n + z^{-n} = (\cos(n\theta) + i\sin(n\theta)) + (\cos(n\theta) - i\sin(n\theta))$ . The imaginary parts cancel out, leaving $2\cos(n\theta)$ .
Let $z = 1 - i\sqrt{3}$ . Find the smallest positive integer $n$ such that $z^n$ is a positive real number.
Answer: 6 First, convert $z = 1 - i\sqrt{3}$ to polar form. The modulus is $r = \sqrt{1^2 + (-\sqrt{3})^2} = \sqrt{1+3} = 2$ . The complex number is in the fourth quadrant, so its argument is $\theta = \arctan(\frac{-\sqrt{3}}{1}) = -\frac{\pi}{3}$ . So, $z = 2e^{-i\pi/3}$ . Using De Moivre's theorem, $z^n = (2e^{-i\pi/3})^n = 2^n e^{-in\pi/3}$ . For $z^n$ to be a real number, its argument must be a multiple of $\pi$ . So, $-\frac{n\pi}{3} = k\pi$ for some integer $k$ , which means $n = -3k$ . For $z^n$ to be a positive real number, its argument must be an even multiple of $\pi$ . So, $-\frac{n\pi}{3} = 2m\pi$ for some integer $m$ , which means $n = -6m$ . We are looking for the smallest positive integer $n$ . Choosing $m=-1$ gives $n = 6$ . Let's check: $z^6 = 2^6 e^{-i(6\pi/3)} = 64e^{-i2\pi} = 64(1) = 64$ , which is a positive real number.
Let the complex number $z$ be represented by $z = \frac{(\sqrt{3}+i)^5}{(1-i)^8}$ . Find the principal argument of $z$ .
Answer: $\frac{5\pi}{6}$ First, convert the numerator and denominator into polar (exponential) form. For the numerator, let $z_1 = \sqrt{3}+i$ . The modulus is $|z_1| = \sqrt{(\sqrt{3})^2+1^2} = 2$ . The argument is $\arg(z_1) = \arctan(1/\sqrt{3}) = \frac{\pi}{6}$ . So, $z_1 = 2e^{i\pi/6}$ . For the denominator, let $z_2 = 1-i$ . The modulus is $|z_2| = \sqrt{1^2+(-1)^2} = \sqrt{2}$ . The argument is $\arg(z_2) = \arctan(-1/1) = -\frac{\pi}{4}$ . So, $z_2 = \sqrt{2}e^{-i\pi/4}$ . Now, use De Moivre's theorem. $z_1^5 = (2e^{i\pi/6})^5 = 32e^{i5\pi/6}$ . $z_2^8 = (\sqrt{2}e^{-i\pi/4})^8 = (\sqrt{2})^8 e^{-i(8\pi/4)} = 16e^{-i2\pi} = 16e^{i0}$ . Then, $z = \frac{32e^{i5\pi/6}}{16e^{i0}} = 2e^{i5\pi/6}$ . The principal argument of $z$ is $\frac{5\pi}{6}$ .

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