jmc

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,$$\begin{array}{rl}{\left(\sqrt{2}\text{cis}\frac{\pi }{4}\right)}^{17}-{\left(\sqrt{2}\text{cis}-\frac{\pi }{4}\right)}^{17}& =2^{17/2}\left(\text{cis}\frac{17\pi }{4}\right)-2^{17/2}\left(\text{cis}-\frac{17\pi }{4}\right)\\ & =2^{17/2}\left[\text{cis}\left(\frac{\pi }{4}\right)-\text{cis}\left(-\frac{\pi }{4}\right)\right]\\ & =2^{17/2}\left(2i\sin \frac{\pi }{4}\right)\\ & =2^{17/2}\cdot 2\cdot 2^{-1/2}i=2^{9}i=\boxed{512}i\end{array}$$