Past exam of the mathematics course of the University of Cambridge / 2012 / ia / Paper 1 / 1C (Vectors and Matrices)

Write $z=1+iy$ with $y\in \mathbb{R}$. In the complex plane, this is the vertical line through $1$. The complex exponential gives $e^z=e(\cos y+i\sin y)$, so its complex modulus is always $e$. Conversely, every point of that modulus has an complex argument $y$ and is obtained in this way. The image is the entire circle centered at zero with radius $e$, traversed repeatedly as $y$ varies; it is not a disk.

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.

Put $z=x+iy$. Comparing imaginary parts gives $-y=1$, so $y=-1$. Comparing real parts then gives $\sqrt{x^2+1}=x$. The left side is positive, so squaring is legitimate only with $x\ge 0$, but even then it would give $x^2+1=x^2$. This contradiction proves there is no solution. Equivalently, with nonzero imaginary part the complex modulus is strictly greater than the real part.