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Re[ (a+bi)^p]

  • Thread starter Gregg
  • Start date
  • #1
459
0

Homework Statement



Evaluate [itex] Re[(a+bi)^p] [/itex]

The Attempt at a Solution



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

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

[itex]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}[/itex]

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:

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31


Why don't you just convert a+bi into polar form to make it easier?
 
  • #3
459
0
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
152
0
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.
 

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