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## 'Complex Sine' printed from http://nrich.maths.org/

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 $$e^{-iz} = 1 -iz + (iz)^2/2! + ...$$