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!

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...

Re: convergence region for complex series

sorry i forgot to put the series sign at the start, between 1 and infinity

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 $\displaystyle \displaystyle \begin{align*} z = 0 - i \end{align*}$.

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 :)

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

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Re: convergence region for complex series