<math>=\frac{e^{-1}(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i)-e(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i)}{2i}=\frac{\frac{\sqrt{2}}{2}(\frac{1}{e}-e)+\frac{\sqrt{2}}{2}(\frac{1}{e}+e)i}{2i}=\frac{\sqrt{2}}{4}(\frac{1}{e}+e)-\frac{\sqrt{2}}{4}(\frac{1}{e}-e)i</math>
====תרגיל====
הוכיחו: <math>\sin(z+w)=\sin z\cos w+\cos z\sin w</math>.
====פתרון=====
נפתח את צד ימין:
<math>\sin z\cos w+\cos z\sin w=\frac{e^{iz}-e^{-iz}}{2i}\cdot \frac{e^{iw}+e^{-iw}}{2}+\frac{e^{iz}+e^{-iz}}{2}\cdot \frac{e^{iw}-e^{-iw}}{2i}=</math>
<math>=\frac{1}{4i}(e^{iz}e^{iw}+e^{iz}e^{-iw}-e^{-iz}e^{iw}-e^{-iz}e^{-iw}+e^{iz}e^{iw}-e^{iz}e^{-iw}+e^{-iz}e^{iw}-e^{-iz}e^{-iw})=\frac{1}{4i}(2e^{iz}e^{iw}-2e^{-iz}e^{-iw})=\frac{e^{i(z+w)}-e^{-i(z+w)}}{2i}=\sin (z+w)</math>