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

The point $1+i$ has modulus $\sqrt{2}$ and argument $\pi /4$ modulo $2\pi$. Taking complex square roots halves the argument and takes the square root of the modulus:The two roots are opposite points in the complex plane, explicitly $\pm [\sqrt{(\sqrt{2}+1)/2}+i\sqrt{(\sqrt{2}-1)/2}]$.

Take the imaginary part of De Moivre's theorem, $(\cos \theta +i\sin \theta {)}^5=\cos 5\theta +i\sin 5\theta$. With $s=\sin \theta$ and ${\cos }^2\theta =1-s^2$,The required equation cancels the fifth powers and becomes $5s(1-4s^2)=0$. Thus $\sin \theta =0,\pm 1/2$, giving all real solutions

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.