Complex numbers — practice questions

First, convert $w = -4\sqrt{3} + 4i$ to exponential form. \ Modulus: $|w| = \sqrt{(-4\sqrt{3})^2 + 4^2} = \sqrt{48 + 16} = \sqrt{64} = 8$. \ Argument: $\arg(w) = \arctan\left(\frac{4}{-4\sqrt{3}}\right) = \arctan\left(-\frac{1}{\sqrt{3}}\right)$. Since the real part is negative and the imaginary part is positive, the number is in the second quadrant. So, $\arg(w) = \pi - \frac{\pi}{6} = \frac{5\pi}{6}$. \ So, $w = 8e^{i(5\pi/6)}$. \ We are solving $z^3 = 8e^{i(5\pi/6)}$. We write this in general form: \ $z^3 = 8e^{i(5\pi/6 + 2k\pi)}$ for $k \in \mathbb{Z}$. \ Taking the cube root: \ $z = (8e^{i(5\pi/6 + 2k\pi)})^{1/3} = 8^{1/3} e^{i\frac{5\pi/6 + 2k\pi}{3}} = 2e^{i(\frac{5\pi}{18} + \frac{2k\pi}{3})}$. \ We find the three distinct roots by taking $k=0, 1, 2$: \ For $k=0$: $z_0 = 2e^{i(5\pi/18)}$. This is in the required range. \ For $k=1$: $z_1 = 2e^{i(5\pi/18 + 2\pi/3)} = 2e^{i(5\pi/18 + 12\pi/18)} = 2e^{i(17\pi/18)}$. This is in the required range. \ For $k=2$: $z_2 = 2e^{i(5\pi/18 + 4\pi/3)} = 2e^{i(5\pi/18 + 24\pi/18)} = 2e^{i(29\pi/18)}$. The argument is outside the range $(-\pi, \pi]$. We adjust it by subtracting $2\pi$: $\frac{29\pi}{18} - 2\pi = \frac{29\pi - 36\pi}{18} = -\frac{7\pi}{18}$. So, $z_2 = 2e^{-i(7\pi/18)}$. \ The solutions are $2e^{i(5\pi/18)}$, $2e^{i(17\pi/18)}$, and $2e^{-i(7\pi/18)}$.

Consider the sum $C+iS$: \ $C+iS = \sum_{r=0}^{n-1} (\cos(r\theta) + i\sin(r\theta)) = \sum_{r=0}^{n-1} (e^{i\theta})^r$. \ This is a geometric series with first term $a = (e^{i\theta})^0 = 1$, common ratio $r = e^{i\theta}$, and $n$ terms. \ The sum is $S_n = \frac{a(1-r^n)}{1-r} = \frac{1(1 - (e^{i\theta})^n)}{1 - e^{i\theta}} = \frac{1 - e^{in\theta}}{1 - e^{i\theta}}$. \ We use the half-angle factorisation trick: $1-e^{i\phi} = e^{i\phi/2}(e^{-i\phi/2} - e^{i\phi/2}) = -2ie^{i\phi/2}\sin(\phi/2)$. \ Applying this to the numerator and denominator: \ Numerator: $1 - e^{in\theta} = -2ie^{in\theta/2}\sin(n\theta/2)$. \ Denominator: $1 - e^{i\theta} = -2ie^{i\theta/2}\sin(\theta/2)$. \ So, $C+iS = \frac{-2ie^{in\theta/2}\sin(n\theta/2)}{-2ie^{i\theta/2}\sin(\theta/2)} = e^{i(n\theta/2 - \theta/2)} \frac{\sin(n\theta/2)}{\sin(\theta/2)} = e^{i(n-1)\theta/2} \frac{\sin(n\theta/2)}{\sin(\theta/2)}$. \ Expanding the exponential term: \ $C+iS = \left(\cos\left(\frac{(n-1)\theta}{2}\right) + i\sin\left(\frac{(n-1)\theta}{2}\right)\right) \frac{\sin(n\theta/2)}{\sin(\theta/2)}$. \ Equating real and imaginary parts: \ $C = \cos\left(\frac{(n-1)\theta}{2}\right) \frac{\sin(n\theta/2)}{\sin(\theta/2)}$. \ $S = \sin\left(\frac{(n-1)\theta}{2}\right) \frac{\sin(n\theta/2)}{\sin(\theta/2)}$.