Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2007 / Paper 1 / Section B / 2

Using the principal complex logarithm, $\mathrm{Log}i=i\pi /2$ and $\mathrm{Log}(-e)=1+i\pi$. Thus the values and their real parts and imaginary parts areFor the final row, first calculate $i^i=e^{-\pi /2}$ and only then raise this positive real number to the power $i$. Treating nested powers as ordinary real powers would ignore the branching of complex exponentiation.

If all branches are requested, $\sqrt{i}$ also has the negative of its principal value, and the complex logarithm of $-e$ is $1+i(\pi +2\pi k)$. Allowing nonprincipal logarithms of the positive base $2$ gives $e^{-2\pi k}{e}^{i\ln 2}$. Finally, all values of $(i^i{)}^i$ are $-i{e}^{-2\pi m}$, for integers $k,m$ as appropriate. These alternatives do not alter the principal-value table.

Euler's formula givesCubing the expression for the sine, and grouping the $e^{3ix}$ and $e^{-3ix}$ terms, givesThe required equation therefore becomes $\sin x(1-4{\sin }^{2}x)=0$. The complete real solution set is

The quadratic formula gives $z=-b\pm \sqrt{b^2-1}$. For $-1\le b\le 1$, write this asEach root of a polynomial has modulus $1$, since $b^2+(1-b^2)=1$. As $b$ increases from $-1$ to $1$, the upper root traverses the upper unit semicircle from $1$ to $-1$, and the lower root traverses the lower semicircle in the opposite angular direction. At $b=-1$ the roots coincide at $1$; at $b=1$ they coincide at $-1$. For intermediate $b$ they are complex conjugates.