Imaginary Numbers to Polar form

[tomeatworld]

1. The problem statement, all variables and given/known data
(1+i)i = reiθ

Find the real values of r and θ.

3. The attempt at a solution
Well, after doing a similar(ish) question I decided taking logs would be a good start:

i loge(1+i) = loger + iθ

From here, I have no idea where to go. Using a power of i is killing me....

[vela]

Start by writing 1+i in polar form.

[tomeatworld]

Right. So I should have:

(\sqrt{2}e(\pi/4) i)i

And from there log?i can't seem to make that get towards a single polar form..

[vela]

Sounds like a good plan. Then you can match the real parts on both sides to each other and similarly with the imaginary parts.

[Mark44]

Right. So I should have:

(\sqrt{2}e(\pi/4) i)i

And from there log?i can't seem to make that get towards a single polar form..
I don't understand how you got this.

1 + i = \sqrt{2}e^{i \pi/4}
\Rightarrow ln(1 + i) = ln(\sqrt{2}e^{i \pi/4}) = ln\sqrt{2} + ln(e^{i \pi/4})

The last term on the right can be simplified.

[tomeatworld]

I got to that as the original question was (1+i)i so I had to put it back into the polar form of (1+i). (unless I'm missing something).

I still can't really see where to go (assuming I've gone the right way).

i (ln \sqrt{2} ei \pi /4)

i (ln \sqrt{2} + ln ei \pi /4)
i (ln \sqrt{2} + i \pi /4 )

and from there just multiply out to get the imaginary and real parts?

[vela]

Yup, because on the RHS, the real part is log r and the imaginary part is \theta.

[tomeatworld]

Ah wow, got it! Thanks a load! Great help!