Complex Sine

Problem | Teachers' Notes | Hint | Solution | Printable page |
Stage: 5 Challenge Level: Challenge Level:3 Challenge Level:3 Challenge Level:3
The formula $$\sin z = {1\over 2\pi}(e^{iz} - e^{-iz})$$ can be verified by showing that the series expansion for $\sin z$, that is $$\sin z = z - z^3/3! + z^5/5! - z^7/7! + ... $$ can be obtained using the series expansions $$e^{iz} = 1 + iz + (iz)^2/2! + ...$$ and $$e^{-iz} = 1 -iz + (iz)^2/2! + ...$$

Published October 2004.