Re[ (a+bi)^p]

[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

[rock.freak667]

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

[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

[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.