Find $\operatorname{Log}(1-i)$ where $\operatorname{Log}z$ is the principal value.

Question

My attempt starts with $\log z=\operatorname{Log}|z| + i\operatorname{Arg}z$. I start with the $\operatorname{Arg}$. We know

$$|1-i|=\sqrt{1^2+1^2}=\sqrt{2}\implies\sin\theta=-\frac{\sqrt{2}}{2}\implies\theta=\frac{7\pi}{4}=-\frac{\pi}{4}$$

So, we have

$$\operatorname{Log}(1-i)=\log z + i(-\frac{\pi}{4})\implies$$ $$\text{Log}(1-i)=\log z - i\frac{\pi}{4}$$

At this point, I'm confused on how we find the $\log z$ part of our statement. I don't think using the definition of $z=e^{\log z}$ is beneficial -- it seems redundant.

Answer 1

Normally, I would not have posted this, but there seems to be confusion among the comments.

I will assume that if $z = re^{i\theta}$,

with $r \in \Bbb{R^+}$

and $-\pi < \theta \leq \pi$ that

$\text{Log}(z) = [\log(r)] + i\theta.$

If $z = (1 - i)$, then $r = \sqrt{2}$ and $\theta = -\pi/4$.

Therefore, $\text{Log}(z) = [\log(\sqrt{2})] + i(-\pi/4).$