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To find the complex numbers z satisfying $\sin(z) = \cos(z)$, can I say: $$\sin(z) = \frac{(e^{iz}-e^{-iz})}{2i}=\frac{(e^{iz}+e^{-iz})}{2}$$

and solve for z? So we then reduce this to $$-e^{-iz} = e^{-iz}$$ but this doesn't look right

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    I don't see how you get from the first equation to the second one.2012-07-13

3 Answers 3