Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 1 / 9 / a

By Euler's formula, $\boxed{e^{2i}=\cos 2+i\sin 2}$, with the angle measured in radians.

Since the arguments of $-1$ are $(2k+1)\pi$, the complex logarithm has values $\boxed{\log (-1)=i(2k+1)\pi }$ for $k\in \mathbb{Z}$. The principal value is $\boxed{i\pi }$.

The modulus of $i$ is one and its arguments are $\pi /2+2k\pi$. Thus $\boxed{\log i=i(\pi /2+2k\pi )}$, with principal value $\boxed{i\pi /2}$.

For the positive real base ten, use its real natural logarithm. Then Euler's formula gives

Using a value of the complex logarithm, $i^i=\exp (i\log i)=\exp [-(\pi /2+2k\pi )]$. Thus every logarithmic-branch value is real and positive:The principal value is $\boxed{e^{-\pi /2}}$.