Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2008 / Paper 1 / 12 / c

Multiplying by $z^2-1$ gives $z^6-1=0$. The sixth roots of unity $1,-1$ do not solve the original equation, since substituting either gives three. The other four do, because the extra factor is nonzero. HenceTheir real parts are $\pm 1/2$ and their imaginary parts are $\pm \sqrt{3}/2$, giving the four corners in the left Argand diagram below.

Writing $z=x+iy$ and comparing real parts and imaginary parts gives $\sin x\cosh y=\cosh 2$ and $\cos x\sinh y=0$. A real solution $y=0$ is impossible since $\cosh 2>1$. Thus $\cos x=0$, and positivity forces $\sin x=1$. Consequently $x=\pi /2+2k\pi$ and $\cosh y=\cosh 2$, givingThe right Argand diagram in the preceding figure shows representative points on the lines $\mathrm{Im}z=\pm 2$; each row continues with spacing $2\pi$.