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

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 $0$.