Past exam of the mathematical methods I course of the University of Oxford / 2019 resit / 1

Since $(1-i{)}^3=-2-2i$, divide and multiply by the complex conjugate of the denominator:Multiplication by $-2-2i$ verifies the result directly.

A complex power requires a choice of complex logarithm. Taking the principal argument of $-e$ to be $+\pi$ gives $\mathrm{Log}(-e)=1+i\pi$, henceThus $a=0$ and $b=e^{-{\pi }^2/2}$ for this principal value. If the power is interpreted as multivalued, the logarithms are $1+i(2k+1)\pi$ and the possible values are $i\exp [-(2k+1){\pi }^2/2]$, $k\in \mathbb{Z}$. Applying an identity for real powers without specifying this branch would lose the ambiguity.