# Homework Help: Engineering Math - Complex Roots and Powers

1. Sep 30, 2011

### Tonik

1. The problem statement, all variables and given/known data
Evaluate the following in x+iy form.
i3+i

2. Relevant equations
i=rei$\Theta$

3. The attempt at a solution
i=rei$\frac{pi}{2}$
i3+i = i3*ii
(-i)(ei2$\frac{pi}{2}$)
=-ie-$\frac{pi}{2}$
=-.2079i

I understand how it all works out except $\frac{pi}{2}$. I can't figure out how they got $\frac{pi}{2}$ in the first place. Any ideas? Thanks in advance.

2. Sep 30, 2011

### Staff: Mentor

My editor won't allow greek letters, so in place of theta I'll use T

You know: eiT=cosT + i.sinT

So let T=Pi/2, so that cosT=0, and we have

ei.Pi/2= i

Q: How do you say in words: rei.theta

I'm wondering what the r is part of?

3. Sep 30, 2011

### HallsofIvy

If $y= i^i$, then $ln(y)= i ln(i)$. Further, i has modulus r= 1 and argument $\theta= \pi/2$ so that $i= e^{i(\pi/2+ 2k\pi)}$ for k any integer ($e^{2k\pi}= 1$ for all integer k so these are all different ways of writing i). Then $ln(i)= i(\pi/2+ 2k\pi)$ and so $ln(y)= -(\pi/2+ 2k\pi)$.

4. Sep 30, 2011

### Tonik

The r comes from converting x+iy (rectangular coordinates) to polar coordinates.
x+iy=r(cos$\Theta$+isin$\Theta$)=rei$\Theta$
As far as how r is used, I'm unsure. None of the problems I have been given/done have used that r at all.

5. Sep 30, 2011

### Tonik

Thank you both for the explanations! You're my heroes.

6. Sep 30, 2011

### Staff: Mentor

I guessed that r was the magnitude, but you confused me by introducing it in (3) for i= and I know i has a magnitude of unity. The square root of (cos2 + sin2) is always 1.

I just wanted to make sure "re" wasn't a combination that I am unfamiliar with.

Well, 5.cosT + 5.i.sinT would equal 5.eiT so that's when it's needed.

7. Sep 30, 2011

### Tonik

Yeah, I think you're probably right. Thanks again! :)