Principal values of complex functions

Question

How do I find the principal value of the following:

$\log(1-\sqrt{2i})$

And hence of

$(1-\sqrt{2i})^i$

Also how do I write

$z=1+i$ in polar form and find its roots?

I find these very confusing so step-by-step answers would be a great help, thanks.

Answer 1

As tpb261 suggested:

$$j = e^{j\pi/2},-1 = e^{j\pi}=e^{-j\pi}$$

$$ 1-\sqrt{2j}=1-2^{1/2}e^{j\pi/4}=1-2^{1/2}\left(\cos(\pi/4)+j \sin(pi/4)\right)$$ $$=1-2^{1/2}\left(2^{-1/2}+j 2^{-1/2}\right)=1-(1+j)=-j= e^{-j\pi/2}$$

Thus

$$\ln(1-\sqrt{2j})=-j\pi/2$$

$$(1-\sqrt{2j})^j=j\exp(\ln((1-\sqrt{2j})))=j\exp(-j\pi/2)=e^{j\pi/2}\exp(-j\pi/2)=1$$

$$\rho e^{j \phi} =z=1+j=2^{1/2}\left(2^{-1/2}+j 2^{-1/2}\right)=2^{1/2}\left(cos(\pi/4)+j\sin(\pi/4)\right)$$ $$=2^{1/2}e^{j\pi/4}$$

Therefore $\rho=2^{1/2}$ and $\phi=\pi/4$.