# Math Help - Power Series Expansions of Complex Valued Functions

1. ## Power Series Expansions of Complex Valued Functions

Here is my problem:
Let sum(a_k*z^k, k=0->infinity)
be the power series expansion of
2/((3-2z)^2).

I need to figure out a closed form expression for a_k.

My first problem is figuring out how to manipulate the function into something I know how to express (geometric series?).

Secondly (although this will probably be easier once I figure out issue number 1) is that I don't see how it will be obvious to express a_k unless you just explicitly get the power series.

any hints?

2. Originally Posted by robeuler
Here is my problem:
Let sum(a_k*z^k, k=0->infinity)
be the power series expansion of
2/((3-2z)^2).

I need to figure out a closed form expression for a_k.

My first problem is figuring out how to manipulate the function into something I know how to express (geometric series?).

Secondly (although this will probably be easier once I figure out issue number 1) is that I don't see how it will be obvious to express a_k unless you just explicitly get the power series.

any hints?
$\int\frac{2}{(3-2z)^2}=\frac{1}{3-2z}=\frac{1}{3}\cdot\frac{1}{1-\frac{2}{3}z}$

So we know that $\forall{z}\backepsilon|z|<1\quad\frac{1}{1-z}=\sum_{n=0}^{\infty}z^n$

So we can ascertain that

$\forall{z}\backepsilon\left|\frac{2}{3}z\right|<1\ quad\frac{1}{3}\cdot\frac{1}{1-\frac{2}{3}z}=\frac{1}{3}\sum_{n=0}^{\infty}\left( \frac{2}{3}z\right)^n$

So on that same interval (due to unifrom convergence, ignore if you dont care or it isnt part of what your learnging) we may differentiate to obtain

\begin{aligned}\left(\frac{1}{3}\frac{1}{1-\frac{2}{3}z}\right)'&=\frac{2}{(3-2z)^2}\\
&=\frac{d}{dz}\frac{1}{3}\sum_{n=0}^{\infty}\left( \frac{2}{3}z\right)^n\\
&=\frac{1}{3}\sum_{n=1}^{\infty}n\left(\frac{2}{3} \right)^nz^{n-1}
\end{aligned}

3. Very nice approach. Uniform continuity is great!

Here is one more I am stuck on, if you have the time:
f(z)=sinz/(pi-z)
g(z)=(z(pi-z))/sinz

I need to find the power series representation of f(z) at pi, and its radius of convergence. Then I need to find the radius of convergence of the power series of g(z) at pi/2.

I can write out sinz as a power series no problem. And again I feel like I should write out the denominator as a geometric series, and then multiply the two series together. I can't figure out how to express it as a geometric series.

4. Originally Posted by robeuler
Very nice approach. Uniform continuity is great!

Here is one more I am stuck on, if you have the time:
f(z)=sinz/(pi-z)
g(z)=(z(pi-z))/sinz

I need to find the power series representation of f(z) at pi, and its radius of convergence. Then I need to find the radius of convergence of the power series of g(z) at pi/2.

I can write out sinz as a power series no problem. And again I feel like I should write out the denominator as a geometric series, and then multiply the two series together. I can't figure out how to express it as a geometric series.
Try writing $\sin(z)=\frac{e^{iz}-e^{-iz}}{2}=\frac{\left(e^i\right)^z-\left(e^i\right)^z}{2}$

5. Originally Posted by Mathstud28
Try writing $\sin(z)=\frac{e^{iz}-e^{-iz}}{2}=\frac{\left(e^i\right)^z-\left(e^i\right)^z}{2}$
I know the expansion for sin(z), or are you suggesting I write it out as you do and then using that to manage the 1/pi-z

6. Originally Posted by robeuler
I know the expansion for sin(z), or are you suggesting I write it out as you do and then using that to manage the 1/pi-z
Well, let me put a disclaimer in here, I am not the best at complex analysis. I am more knowledgable of real valued series, but what I am saying is consider that $e^{iz}$ is geometric.

7. Originally Posted by Mathstud28
Try writing $\sin(z)=\frac{e^{iz}-e^{-iz}}{2}=\frac{\left(e^i\right)^z-\left(e^i\right)^z}{2}$
That is not true.

Just consider this fallacy,
$1 = 1^{\pi} = (e^{2\pi i})^{\pi} = e^{2\pi^2 i} = \cos (2\pi ^2) + i\sin (2\pi^2)$

8. Originally Posted by ThePerfectHacker
That is not true.

Just consider this fallacy,
$1 = 1^{\pi} = (e^{2\pi i})^{\pi} = e^{2\pi^2 i} = \cos (2\pi ^2) + i\sin (2\pi^2)$
Thank you, what I said is only true for real values isn't it? Im sorry Robeuler, I had no buisness responding to this post. It is not something I am well versed in. And thank you TPH for pointing that out. I will try to be better about that.

9. Originally Posted by Mathstud28
Thank you, what I said is only true for real values isn't it? Im sorry Robeuler, I had no buisness responding to this post. It is not something I am well versed in. And thank you TPH for pointing that out. I will try to be better about that.
If $a,b,c$ are positive real numbers then we have the exponents rules:
1. $a^ba^c = a^{b+c}$
2. $(a^b)^c = a^{bc}$

If $a,b,c$ are (non-zero) complex numbers then we may ask if:
1. $a^ba^c = a^{b+c}$
2. $(a^b)^c = a^{bc}$

It turns out that #1 is always true. But #2 needs to be used with a lot of care. One way when #2 always works is when $c \in \mathbb{Z}$. You certainly seen that $(e^{i\theta})^n = e^{ni\theta}$. Well, that is another special case when $c$ is an integer.

10. Originally Posted by Mathstud28
Thank you, what I said is only true for real values isn't it? Im sorry Robeuler, I had no buisness responding to this post. It is not something I am well versed in. And thank you TPH for pointing that out. I will try to be better about that.