= Solution
Take $r>0$ and select a <complex argument> $\theta$ for $z$. To specify the <complex exponentiation>, use $z^{1+i}=\exp((1+i)(\ln r+i\theta))$. Expanding its exponent gives
$$
z^{1+i}=e^{\ln r-\theta}e^{i(\theta+\ln r)}.
$$
Its <complex logarithm> therefore has values
$$
\boxed{\Re f=\ln r-\theta,\qquad\Im f=\theta+\ln r+2\pi k,\quad k\in\mathbb Z.}
$$
The continuous logarithm obtained by retaining the displayed exponent uses $k=0$. If the outer logarithm is the <principal logarithm>, choose $k$ to put the <imaginary part> in $(-\pi,\pi]$ instead; the <real part> is unchanged. In particular one cannot generally assert $\operatorname{Log}(e^w)=w$ without checking its <imaginary part>.
The condition on the <real part> gives $\ln r=\theta$, so the required <logarithmic spiral> has <polar coordinates>
$$
\boxed{r=e^\theta,\qquad z(\theta)=e^\theta(\cos\theta+i\sin\theta).}
$$
As $\theta$ increases, it runs from the negative <real axis> near radius $e^{-\pi}$ through $-ie^{-\pi/2}$, $1$ and $ie^{\pi/2}$ to the negative <real axis> at radius $e^\pi$. The <angle> increases by one complete turn. Using the conventional <principal argument> $-\pi<\theta\le\pi$, the inner negative-axis endpoint is excluded and the outer endpoint is included. The convention $-\pi\le\theta<\pi$ reverses those endpoint choices. If both limiting arguments are admitted by the paper's wording, both limiting endpoints are drawn; the interior curve is the same. The origin is never included because the <complex logarithm> is undefined there.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2019/paper-1-2019-q11-logarithmic-spiral.png]
{title=One turn of the logarithmic spiral, with an enlarged view near the origin}
Back to article page
