Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2019 / Paper 1 / 11 / b / Solution

The complex conjugate is $z^{\ast }=x-iy$. From the complex sine and cosine identities,Divide these expressions and multiply by the complex conjugate of the denominator. Its squared modulus is ${\cos }^{2}x{\cosh }^{2}y+{\sin }^{2}x{\sinh }^{2}y=(\cos 2x+\cosh 2y)/2$. The numerator simplifies using ${\cosh }^{2}y-{\sinh }^{2}y=1$ and the double-angle formulas. ThusThese expressions apply except at the poles $y=0$, $x=\pi /2+k\pi$ with integer $k$, where the original tangent function is undefined.