Prove all values of $(1-i)^{\sqrt2i}$ lie on a straight line in the complex field

Question

I'm trying to prove that all of the values of $(1-i)^{\sqrt2i}$ lie on a straight line. Can someone explain what I should be doing and point me in the right direction to get started? That looks like a single value to me, not a function with multiple values. How can I look at that expression as something with multiple values?

Answer 1

Note that in general for a complex $z=|z|e^{i\theta}$ you have that:

$$ln(z)=ln|z|+i\theta$$

So for example if you want to calculate $x=i^i$: $$ln(x)=i \left( ln(1)+i\left(\frac{\pi}{2}+2k\pi \right) \right)=-\frac{\pi}{2}+2k\pi,\space k\in Z $$ Then, $$x=exp \left({-\frac{\pi}{2}+2k\pi}\right),\space k\in Z$$

So see that there are infinitely many answers, beacuse there are infinitely many angles, so that happens when you raise a complex number to another complex number. Note that if you google $i^i$, google will tell you that the answer is $0.207879...$, and this is what you get for $k=0$, so the answer by deafult is $e^{-\pi/2}$, even though there are infinitely more solutions.

Answer 2

Hint: The expression $z^w$ is interpreted as $e^{w\log z}$. Note that $\log z$ has an infinite number of values, all differing by integral multiples of $2\pi i$.