Holomorphic Branches

Let C denote spiral given by r= 2e^θ (in polars) . There is a unique holomorphic branch of log on ℂ\C s.t log(1)= 0.

For this branch determine:

log(i), log(3), log(-1), log(1000), log(-1000), log(2000)


I'm struggling to understand the intuition here. I know about how holomorphic branches of L(z) can can be found by restricting to say ℂ\[-inf, 0] etc, and we do this so the value of L(z) can be single valued. However, I'm unsure how to define L(z) in the case of a spiral. I have an idea but its just a guess really:

Say we define L(z) = log(|z|) + i*arg(z) + 2n*pi, where n is the number of times the spiral has wrapped around the origin by the time we get to the complex number we need (say for 1000, the spiral has wrapped around 0 times, but for 2000 its wrapped around once) (I'm sort of thinking of this in a 3d plane)

However, I've never seen an example with spirals (or infact any other curve that isn't a line in the complex plane) so I'm not sure if this is anywhere near. Any help appreciated

Still at A level and have only watched videos regarding this topic and wiki...

However,

Draw the spiral by sampling $\theta$ at least every $\frac{2\pi}{3}$, between two arms of the spiral there is a curved strip of the complex plane. Following $\arg(z)$ by continuity (starting from $\arg(1) = 0)$ you get a branch of $\log(z)$ on $\mathbb{C}/C$

Hope this helps

Say we define L(z) = log(|z|) + i*arg(z) + 2n*pi, where n is the number of times the spiral has wrapped around the origin by the time we get to the complex number we need (say for 1000, the spiral has wrapped around 0 times, but for 2000 its wrapped around once) (I'm sort of thinking of this in a 3d plane and 'moving up a level')

You have the right idea.

log(1)=0. And this branch will agree with the standard lnx in the range

exp(-2pi) < x < exp(2pi) of the real axis

To the right of exp(2pi) and as far as exp(4pi) the function is lnx + 2 pi i as we are one level up.

Across the curve C there will always be a 2 pi i discontinuity.