Mock AIME 1 2007-2008 Problems/Problem 5

From AoPSWiki

Problem 5

Let $S = (1+i)^{17} - (1-i)^{17}$ , where $i=\sqrt{-1}$ . Find $|S|$ .

Solution

Rewriting the complex numbers in polar notation form, $1+i = \sqrt{2}\,\text{cis}\,\frac{\pi}{4}$ and $1-i = \sqrt{2}\,\text{cis}\,-\frac{\pi}{4}$ , where $\text{cis}\,\theta = \cos \theta + i\sin \theta$ . By De Moivre's Theorem, $\begin{align*}\left(\sqrt{2}\,\text{cis}\,\frac{\pi}{4}\right)^{17} - \left(\sqrt{2}\,\text{cis}\,-\frac{\pi}{4}\right)^{17} &= 2^{17/2}\,\left(\text{cis}\,\frac{17\pi}{4}\right) - 2^{17/2}\,\left(\text{cis}\,-\frac{17\pi}{4}\right) \\&= 2^{17/2}\left[\text{cis}\left(\frac{\pi}{4}\right) - \text{cis}\left(-\frac{\pi}{4}\right)\right] \\&= 2^{17/2}\left(2i\sin \frac{\pi}{4}\right) \\&= 2^{17/2} \cdot 2 \cdot 2^{-1/2}i = 2^9i = \boxed{512}\,i\end{align*}$

Mock AIME 1 2007-2008 (Problems, Source)
Preceded by Problem 4	Followed by Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15

Personal tools

Navigation

Search

Toolbox

Mock AIME 1 2007-2008 Problems/Problem 5

From AoPSWiki

Problem 5

Solution

See also