# convergence region for complex series

• Apr 12th 2012, 07:26 AM
cat91
convergence region for complex series
Hi, recently got given this question, asked around as well as googled it but can't seem to figure how to do it. We've never done a question with both z and n in it.

Find the region of convergence for this complex series and draw the region:

[(i + z)^(2n - 1)] / [2^(2n + 1)]

thank you!
• Apr 12th 2012, 07:52 AM
Prove It
Re: convergence region for complex series
Quote:

Originally Posted by cat91
Hi, recently got given this question, asked around as well as googled it but can't seem to figure how to do it. We've never done a question with both z and n in it.

Find the region of convergence for this complex series and draw the region:

[(i + z)^(2n - 1)] / [2^(2n + 1)]

thank you!

What you have written is not a series...
• Apr 12th 2012, 08:00 AM
cat91
Re: convergence region for complex series
sorry i forgot to put the series sign at the start, between 1 and infinity
• Apr 12th 2012, 08:49 AM
Prove It
Re: convergence region for complex series
Quote:

Originally Posted by cat91
Hi, recently got given this question, asked around as well as googled it but can't seem to figure how to do it. We've never done a question with both z and n in it.

Find the region of convergence for this complex series and draw the region:

[(i + z)^(2n - 1)] / [2^(2n + 1)]

thank you!

\displaystyle \displaystyle \begin{align*} \sum_{n = 1}^{\infty}\frac{(i + z)^{2n-1}}{2^{2n+1}} &= \sum_{n = 1}^{\infty}\frac{(i+z)^{2n-1}}{2^{2}2^{2n-1}} \\ &= \frac{1}{4}\sum_{n = 1}^{\infty}\left(\frac{i + z}{2}\right)^{2n - 1} \end{align*}

The series is convergent where \displaystyle \displaystyle \begin{align*} \left|\frac{a_{n+1}}{a_n}\right| < 1 \end{align*}, so

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

So the series is convergent for all \displaystyle \displaystyle \begin{align*} z \end{align*} inside the circle of radius 2 units centred at \displaystyle \displaystyle \begin{align*} z = 0 - i \end{align*}.
• Apr 12th 2012, 10:11 AM
cat91
Re: convergence region for complex series
ah i never thought to make 2^(2n+1) = 2^(2).2^(2n-1)
that makes much more sense now thank you :)
• Apr 14th 2012, 11:00 AM
uopdon
Re: convergence region for complex series
Dear cat91 ~ thankyou for starting this thread and big up for your initative. So we're looking at the Ratio Test, and delving deeper into the test and finding the region of convergence aswell. 'Simple like that!' will be pleased. See you in class xx

Dear Prove It, thankyou for your excellent answer and invaluable help. Would it be appropriate to ask you how you embed mathType objects into the threads?

regards, Don
• Apr 14th 2012, 11:07 AM
uopdon
Re: convergence region for complex series