Imaginary Numbers to Polar form

[tomeatworld]

Homework Statement

(1+i)ⁱ = re^iθ

Find the real values of r and θ.

The Attempt at a Solution

Well, after doing a similar(ish) question I decided taking logs would be a good start:

i log_e(1+i) = log_er + 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} e^{i \pi/4})

i (ln \sqrt{2} + ln e^{i \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!