# Re[ (a+bi)^p]

1. May 25, 2009

### Gregg

1. The problem statement, all variables and given/known data

Evaluate $Re[(a+bi)^p]$

3. The attempt at a solution

$(a+bi)^p =\sum _{k=0}^p \left( \begin{array}{c} p \\ k \end{array} \right) a^{p-k} (\text{bi})^k$

$Re[(a+bi)^p] =\sum _{k=0}^p \left( \begin{array}{c} p \\ k \end{array} \right) a^{p-k} (\text{bi})^k$

$Re[\displaystyle \sum _{k=0}^p \text{bi}^k a^{p-k} \left( \begin{array}{c} p \\ k \end{array} \right)] = \sum _{k=0}^{p/2} \left( \begin{array}{c} p \\ 2k \end{array} \right) a^{p-2k} (\text{bi})^{2k}$

I just thought that for each even power of bi that that part will be real. The answer is completely different though. Just confused.

http://www.exampleproblems.com/wiki/index.php/CV8

Last edited: May 25, 2009
2. May 25, 2009

### rock.freak667

Re: Re[I9 (a+bi)^p]

Why don't you just convert a+bi into polar form to make it easier?

3. May 25, 2009

### Gregg

Oh right yeah, that makes it very easy to find. In the solution the modulus isn't included though? I thought it would be that multiplied by the modulus of the a+bi bit

4. May 25, 2009

### Chrisas

For what it's worth, I think that modulus^p should be out in front of that cosine in the link. Unless the modulus is specified to be of 1 someplace.