Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2010 / Paper 1 / 14 / a

Multiply by the complex conjugate of the denominator:Thus

The logarithm's argument is $\sqrt{3}/2+i/2=e^{i\pi /6}$, with complex modulus one. The principal complex logarithm is therefore $\ln 1+i\pi /6=i\pi /6$, givingIf $\ln z$ denotes all logarithm values rather than the principal branch, they are $i(\pi /6+2\pi k)$ for $k\in \mathbb{Z}$: the real part remains zero and the imaginary part changes by integer multiples of $2\pi$. The principal value is the conventional single-valued answer.

Integer powers of $i$ repeat with period four. Thus $i^{-5}=i^{-4}{i}^{-1}=1/i=-i$, andThis negative integer power is single-valued and does not require a logarithm branch.

The number being squared has polar form of a complex number $e^{2\pi i/3}$. Squaring doubles its argument, giving $e^{4\pi i/3}=-1/2-i\sqrt{3}/2$. Hence