# Imaginary Numbers to Polar form

1. Jan 6, 2010

### tomeatworld

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

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....

2. Jan 6, 2010

### vela

Staff Emeritus
Start by writing 1+i in polar form.

3. Jan 6, 2010

### 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..

Last edited: Jan 6, 2010
4. Jan 6, 2010

### vela

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

5. Jan 6, 2010

### Staff: Mentor

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.

6. Jan 6, 2010

### 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?

7. Jan 6, 2010

### vela

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

8. Jan 6, 2010

### tomeatworld

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