Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 1 / 1C / b

For $z=x+iy$, the complex exponential has complex modulus $e^x$. Equality to the positive real number $e$ therefore forces $x=1$ and $\cos y+i\sin y=1$. The latter holds precisely when $y$ is an integer multiple of $2\pi$. Hence all solutions areSubstitution verifies every listed value; the modulus and angle arguments exclude any others.

Let $w$ denote the value of the complex logarithm used in the equation. Factoring $w^2+{\pi }^2/4$ gives $w=i\pi /2$ or $w=-i\pi /2$. Since $z=e^w$, the complete answer isThe principal complex logarithm takes exactly these values at $i$ and $-i$, so both are solutions with that convention. If the complex logarithm is interpreted as multivalued, the same answer holds in the sense that a logarithm value satisfies the equation. The equation is not an assertion that every logarithm value of these numbers has the stated square.