GRE Problems - Variants

Solution This problem relies on the polar represenation of complex numbers. This in turn is strictly related to the polar coordinates studied in the Calculus course. Given a point \((x,y)\) in \({\mathbb R}^2\), we define \(r, \theta\) by \begin{eqnarray*}x&=&r\cos \theta \\ y&=&r\sin \theta\end{eqnarray*} This actually defines \(x\) and \(y\) in terms of \(r\) and \(\theta\). To find \(r\) and \(\theta \) from \(x\) and \(y\), we use the formulas \begin{eqnarray*}r&=&\sqrt{x^2+y^2} \\ \tan \theta&=&\frac{y}{x}.\end{eqnarray*} We can use the \(\tan^{-1}\) function to solve for \(\theta\), but with care. Remember that the range of the \(\tan^{-1}\) is \((-\pi/2,\pi/2)\). That means only quadrants I and IV. If our point is in quadrant II or III, we need to add \(\pi\) to the output given by \(\tan^{-1}\). So for example suppose \((x,y)=(-3,4)\), in quadrant II. Using the \(\tan^{-1}\) on a calculator, we find \(\tan^{-1}(-4/3)\approx -53.13^\circ\). But the correct value for \(\theta\) is \(-53.13+180=126.87^\circ\).
Complex numbers in Cartesian coordinates are of the form \(z=x+iy\), where \(i^2 = -1\). If we substitute \(x=r\cos\theta\) and \(y=r\sin \theta\), we find \[z=r(\cos \theta + i \sin \theta).\] At this point we we use the all-important Euler's formula relating the trigonometric and exponential functions: \[e^{i\theta} = \cos\theta + i\sin \theta.\] So we can write a complex number \(z\) in polar coordinates as \[z= re^{i\theta}.\] This is of course very handy when taking powers, because \(\left(e^{i\theta}\right)^n=e^{in\theta}\), which means that all we need to do is multiply the angle \(\theta \) by \(n\). For a comparison, \((x+iy)^n\) requires instead that we expand a binomial raised to the \(n\)-th power, a much more complicated operation.
So the first step in solving the problem is converting the given complex number to polar coordinates. We have \[x=-1, \hspace{2ex} y=\sqrt{3}.\] So \[r=\sqrt{(-1)^2+(\sqrt{3})^2}=\sqrt{1+3}=2,\] and \[\tan(\theta)=\frac{\sqrt{3}}{-1}=-\sqrt{3}.\] Since \(\tan^{-1}(-\sqrt{3})= -\pi/3\) is in quadrant IV, while the given angle is in quadrant II, we add \(\pi\) and find \(\theta =2\pi/3\). So we have found \[-1+i\sqrt{3} = 2e^{2\pi i/3}.\] Raising this to the 13-th power gives \[ (-1+i\sqrt{3})^{13} = \left(2e^{2\pi i/3}\right)^{13}=2^{13} e^{26\pi i/3}.\] Since \(26\pi/3 = 24\pi/3 + 2\pi/3 = 8\pi+2\pi/3\), we replace \(26\pi/3\) with \(2\pi/3\) and find (using Euler's formula again) \[(-1+i\sqrt{3})^{13}=8192 e^{2\pi i/3}=8192 \left(\cos\frac{2\pi}{3}+ i \sin\frac{2\pi}{3}\right) =8192\left(-\frac{1}{2}+i\frac{\sqrt{3}}{2}\right) = 4096i\sqrt{3}-4096.\]