Understanding the Real Component of a Complex Number Raised to a Power

[Gregg]

Homework Statement

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

The Attempt at a Solution

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

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

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

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.