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

Squaring the modulus and doubling the complex argument givesThis uses Euler's formula $e^{i\theta }=\cos \theta +i\sin \theta$.

Multiply numerator and denominator by the complex conjugate of the denominator:The denominator is the squared modulus $|1-i{|}^2=2$.

The complex number $\sqrt{3}+i$ has modulus two and principal argument $\pi /6$. Therefore its principal complex logarithm isIf the source's logarithm means the full multivalued complex logarithm, rather than its principal value, the complete answer is $\ln 2+i(\pi /6+2\pi n)$ with $n\in \mathbb{Z}$. A branch convention is necessary to obtain a unique imaginary part.

Use the principal complex logarithm at each stage of the nested complex exponentiation. Since $\mathrm{Log}i=i\pi /2$,This inner value is positive real, so its principal complex logarithm is $-\pi /2$. ConsequentlyOne must retain the parentheses: complex exponentiation is not associative. If all complex logarithm branches are allowed, the inner values are $e^{-\pi /2-2\pi k}$. Taking the outer logarithm with an independent integer branch index $\ell$ yields $\exp [-i\pi /2-2\pi ik-2\pi \ell ]=-i{e}^{-2\pi \ell }$. Thus the complete multivalued set is $\{-i{e}^{-2\pi \ell }:\ell \in \mathbb{Z}\}$, and the boxed answer selects its principal evaluation.