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 / ii / Solution

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 $2\pi$.