Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / 12

Multiply numerator and denominator by the complex conjugate of the denominator:Squaring gives $(-1-3i{)}^2/4=(-8+6i)/4$. Thus the real part and imaginary part are

In polar form of a complex number, $1+i=\sqrt{2}{e}^{i\pi /4}$. By De Moivre's theorem,

The complex sine function identity $\sin (u+iv)=\sin u\cosh v+i\cos u\sinh v$ follows from the exponential definition of sine. With $u=\pi /2$, its imaginary term vanishes, whileConsequently the real part is $5/4$ and the imaginary part is $0$.

Write a square root as $u+iv$, with $u,v$ real. Equating the real part and imaginary part of its square gives $u^2-v^2=-1$ and $2uv=1$. Taking complex modulus also gives $u^2+v^2=\sqrt{2}$. ThereforeSince $uv>0$, the signs agree. The two roots areThe $\pm$ changes both signs together, rather than independently.

Set $z=e^{i\theta }$, so $2i\sin \theta =z-z^{-1}$. The binomial theorem givesUse $z^k-z^{-k}=2i\sin (k\theta )$ and $(2i{)}^5=32i$ to obtainThis is a finite Fourier sine series, with only odd harmonics.

Using the exponential definition of hyperbolic cosine, multiply $e^z+e^{-z}=-2$ by the nonzero quantity $e^z$. This yields $(e^z+1{)}^2=0$, hence $e^z=-1$.

If $z=x+iy$, its complex modulus requires $e^x=1$, so $x=0$; its phase then requires $y=(2k+1)\pi$. Thus all complex solutions areThere are no real solutions, since real $\cosh x\ge 1$.