Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2005 / Paper 2 / 5 / c / Solution

Use the exponential definitions of complex sine function and complex cosine. Multiplying the equation by $2i$ gives $e^{iz}-e^{-iz}=-2(e^{iz}+e^{-iz})$, hence $3e^{iz}+e^{-iz}=0$. Since exponentials never vanish, this is equivalent to $e^{2iz}=-1/3$. All branches of the complex logarithm giveThe derivation does not divide by $\cos z$, so it introduces no lost zero-denominator cases.