Past exam of the mathematics course of the University of Cambridge / 2011 / ia / Paper 1 / 1C (Vectors and Matrices)

For a nonzero complex number, write $z=r{e}^{i\theta }$ with $r=|z|>0$ and principal complex argument $-\pi <\theta \le \pi$. The principal complex logarithm and corresponding complex exponentiation areThe value at zero is not defined by this formula. The choice of complex argument matters: these are single-valued principal powers, rather than all values of a multivalued complex logarithm.

For the requested sketch, $(1+i)\mathrm{Log}z=(\log r-\theta )+i(\log r+\theta )$. Its complex exponential has complex modulus $e^{\log r-\theta }$, soThis is one turn of a logarithmic spiral, with increasing radius as the polar angle increases. It passes through $1$ at $\theta =0$, approaches the excluded endpoint $-e^{-\pi }$ from below the negative real axis, and ends at the included endpoint $-e^{\pi }$ from above that axis. The endpoint distinction follows from the chosen principal complex argument.

The principal complex exponentiation gives $z^i=e^{-\theta }{e}^{i\log r}$. Equality to one requires its complex modulus to be one, hence $\theta =0$, and then its complex argument requires $\log r\in 2\pi \mathbb{Z}$. Thus all solutions are positive real, namelySubstitution verifies every value, and the modulus condition excludes every nonreal value.

Away from the negative real axis, the complex conjugate has principal complex argument $-\theta$. ConsequentlyTaking the complex modulus gives $\cosh \theta =1$, hence $\theta =0$. The remaining equation is $e^{i\log r}=i$, soOn the negative real axis, both $z$ and $\overline{z}$ have principal complex argument $\pi$, so the left-hand side has modulus $2e^{-\pi }$, not two. There are no additional solutions there. The printed second term is the principal power of $\overline{z}$, not the complex conjugate of $z^i$.