# Sum of complex power series

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• Apr 9th 2013, 04:12 PM
gralla55
Sum of complex power series
I've attached a picture on the problem and my progress so far.

What confuses me is the i in the denominator. I'm guessing the limit is just 1 / (1 - i). If so, |z|(1 - i) > 1 gives me the radius of convergence. This does not look like a geometric series, so I'm not sure how to find the sum...
• Apr 9th 2013, 09:18 PM
hollywood
Re: Sum of complex power series
The expressions n and n+1 are both real, so you can bring them outside of the absolute value signs.

As for finding the sum, if you differentiate the geometric series $\sum_{n=0}^\infty x^n$ term-by-term, you get $\sum_{n=1}^\infty nx^{n-1}$. I think that will help.

- Hollywood
• Apr 10th 2013, 10:08 AM
gralla55
Re: Sum of complex power series
Thank you! So, taking the derivative of this series won't get me anywhere, so I'm guessing I have to take the anti-derivative, find the sum of that, and finally take the derivative of the result.

However, when taking the antiderivative, I just got stuck again. Maybe I did the antiderivative wrong, or there's something more that could be done to make the expression into a geometric series. I've attached a picture of the antiderivative I got.
• Apr 10th 2013, 05:38 PM
Prove It
Re: Sum of complex power series
Quote:

Originally Posted by gralla55
I've attached a picture on the problem and my progress so far.

What confuses me is the i in the denominator. I'm guessing the limit is just 1 / (1 - i). If so, |z|(1 - i) > 1 gives me the radius of convergence. This does not look like a geometric series, so I'm not sure how to find the sum...

$\displaystyle f(z) &= \sum_{n = 0}^{\infty}{\frac{n\,z^n}{(1-i)^n}}$

By the ratio test, we know that the sum will be convergent when $\displaystyle \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| < 1$. So that means we have to have

\displaystyle \begin{align*} \lim_{n \to \infty} \left| \frac{\frac{\left( n + 1 \right) z^{n+1}}{(1-i)^{n+1}}}{\frac{n\,z^n}{(1-i)^n}} \right| &< 1 \\ \lim_{n \to \infty} \left| \frac{\left( n + 1 \right) z}{n\left( 1 - i \right) } \right| &< 1 \\ \frac{|z|}{|1 - i|} \lim_{n \to \infty} \left| \frac{n+1}{n} \right| &< 1 \\ \frac{|z|}{\sqrt{2}} \cdot 1 &< 1 \\ |z| &< \sqrt{2} \end{align*}

So the radius of convergence is $\displaystyle \sqrt{2}$.
• Apr 10th 2013, 10:53 PM
hollywood
Re: Sum of complex power series
To get the sum, you need to bring a factor of z out:

$f(z) = \sum_{n = 0}^{\infty}{\frac{n\,z^n}{(1-i)^n}} = \sum_{n = 1}^{\infty}{\frac{n\,z^n}{(1-i)^n}} = z\sum_{n = 1}^{\infty}{\frac{n\,z^{n-1}}{(1-i)^n}}$

Now you "un-differentiate" - I made up that word because I don't really want to say "integral" or "antiderivative":

$f(z) = z\frac{d}{dz}\sum_{n = 1}^{\infty}{\frac{z^n}{(1-i)^n}}$

So now you have a geometric series.

- Hollywood
• Apr 11th 2013, 01:10 AM
Prove It
Re: Sum of complex power series
Quote:

Originally Posted by hollywood
To get the sum, you need to bring a factor of z out:

$f(z) = \sum_{n = 0}^{\infty}{\frac{n\,z^n}{(1-i)^n}} = \sum_{n = 1}^{\infty}{\frac{n\,z^n}{(1-i)^n}} = z\sum_{n = 1}^{\infty}{\frac{n\,z^{n-1}}{(1-i)^n}}$

Now you "un-differentiate" - I made up that word because I don't really want to say "integral" or "antiderivative":

$f(z) = z\frac{d}{dz}\sum_{n = 1}^{\infty}{\frac{z^n}{(1-i)^n}}$

So now you have a geometric series.

- Hollywood

What's wrong with the words "antiderivative" and "antidifferentiate"?
• Apr 11th 2013, 04:06 PM
hollywood
Re: Sum of complex power series
Nothing at all wrong with the words - nothing wrong with "integrate" or "integral" either. I was just thinking that this is not the type of problem where we typically use those words.

Upon further reflection, however, perhaps "antiderivative" is actually the perfect word for this situation....

- Hollywood