Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2009 / Paper 1 / 14

The values of the complex logarithm at $-1$ are $i(2m+1)\pi$, $m\in \mathbb{Z}$. For any such value and any integer $n$,This is an integer multiple of $\pi$, so its sine is zero. In particular the result holds with the principal value $\log (-1)=i\pi$. The argument does not require treating the multivalued logarithm as a single globally defined function.

Using the principal complex square root, $\sqrt{i}=e^{i\pi /4}$. Its two square roots differ by a sign, givingEquivalently their arguments are $\pi /8$ and $9\pi /8$ modulo $2\pi$. If instead the right-hand square root is allowed to take both values, the union has four elements $e^{i(\pi /8+k\pi /2)}$, $k=0,1,2,3$, satisfying $z^4=i$. Only the two boxed values square to the principal right-hand side.

Substitute the exponential definitions of hyperbolic cosine and hyperbolic sine. The cross terms cancel:This proves the identity for real arguments, and the same algebra proves it for complex arguments.

The exponential definitions giveThus the double-angle identity follows directly, without any restriction beyond the domain of the exponential.