# [SOLVED] Complex Taylor, Maclaurin and Laurent series

• November 17th 2008, 12:17 PM
Spec
[SOLVED] Complex Taylor, Maclaurin and Laurent series
1) Find the Taylor series around the point 2i and specify the radius of convergence:

(a) $f(z)=\frac{1}{z}$
(b) $f(z)=log(z)$

(a): $f^{(n)}(z)=\frac{(-1)^n n!}{z^{n+1}}$ and $c_n= \frac{f^{(n)}(z_0)}{n!}=\frac{(-1)^n}{{z_0}^{n+1}}$ where $z_0 = 2i$

So the Taylor series is: $\sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^{n+1}}(z-2i)^n=\frac{1}{2i}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^n}(z-2i)^n=\frac{1}{2i}\sum_{n=0}^{\infty}(-1)^{n+1}\left(\frac{i}{2}\right)^n(z-2i)^n$

My textbook has a slightly different answer though. And the radius of convergence? Do I need to use the ratio test, or is there some simpler way to deal with it?

2) Find the first three nonvanishing terms for the Maclaurin series of: $f(z)=\frac{(1-z)^{1/2}}{1+z^2}$

3) Determine the Laurent series for $\frac{1}{z+z^2}$, $|z+1/2|> 1/2$

I've solved this for different areas, but I'm having a little trouble with this one.
• November 17th 2008, 01:16 PM
Opalg
Quote:

Originally Posted by Spec
1) Find the Taylor series around the point 2i and specify the radius of convergence:

(a) $f(z)=\frac{1}{z}$
(b) $f(z)=log(z)$

(a): $f^{(n)}(z)=\frac{(-1)^n n!}{z^{n+1}}$ and $c_n= \frac{f^{(n)}(z_0)}{n!}=\frac{(-1)^n}{{z_0}^{n+1}}$ where $z_0 = 2i$

So the Taylor series is: $\sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^{n+1}}(z-2i)^n =\frac{1}{2i}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^n}(z-2i)^n$

Correct as far as there,...

Quote:

Originally Posted by Spec
$= \frac{1}{2i}\sum_{n=0}^{\infty}(-1)^{n+1}\left(\frac{i}{2}\right)^n(z-2i)^n$

... but that step is wrong. In fact, $\frac{-1}i = i$, so $\frac{(-1)^n}{(2i)^n} = \left(\frac{i}{2}\right)^n$. The $(-1)^{n+1}$ ought not to be there at all.

Quote:

Originally Posted by Spec
And the radius of convergence? Do I need to use the ratio test, or is there some simpler way to deal with it?

The ratio test is always good for finding a radius of convergence. In this case, it's a bit quicker if you notice that this is a geometric series.
• November 17th 2008, 01:32 PM
shawsend
Quote:

Originally Posted by Spec
1) Find the Taylor series around the point 2i and specify the radius of convergence:

(a) $f(z)=\frac{1}{z}$
(b) $f(z)=log(z)$

(a): $f^{(n)}(z)=\frac{(-1)^n n!}{z^{n+1}}$ and $c_n= \frac{f^{(n)}(z_0)}{n!}=\frac{(-1)^n}{{z_0}^{n+1}}$ where $z_0 = 2i$

So the Taylor series is: $\sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^{n+1}}(z-2i)^n=\frac{1}{2i}\sum_{n=0}^{\infty}\frac{(-1)^n}{(2i)^n}(z-2i)^n=\frac{1}{2i}\sum_{n=0}^{\infty}(-1)^{n+1}\left(\frac{i}{2}\right)^n(z-2i)^n$

My textbook has a slightly different answer though. And the radius of convergence? Do I need to use the ratio test, or is there some simpler way to deal with it?

2) Find the first three nonvanishing terms for the Maclaurin series of: $f(z)=\frac{(1-z)^{1/2}}{1+z^2}$

3) Determine the Laurent series for $\frac{1}{z+z^2}$, $|z+1/2|> 1/2$

I've solved this for different areas, but I'm having a little trouble with this one.

Should not the radius of convergence of a Taylor series simply be the distance to the nearest singularity? In that case, the radius of convergence for (1a) would be 2? I would think the same with (1b) except the series will converge to an analytic sheet of $log(z)$

Also, (3) sounds not right. Should it be $|z+1/2|<1/2$ or no?
 ok, I understand now. I' wrong. Sorry. . .
• November 17th 2008, 01:51 PM
Spec
I solved (3) myself by using the substitution $w=z+1/2$

Then you have:

$\frac{1}{w^2-1/4}=\frac{1}{w^2}\frac{1}{1-\frac{1/4}{w^2}}=\frac{1}{w^2}\sum_{n=0}^{\infty}\frac{1/4}{w^2}^n = ...$

What about the Taylor series for $log z$ around the point $2i$?
• November 17th 2008, 11:53 PM
Opalg
Quote:

Originally Posted by Spec
What about the Taylor series for $\log z$ around the point $2i$?

$\log z = \log(2i + z - 2i) = \log(2i) + \log\bigl(1 + \tfrac{z-2i}{2i}\bigr)$. Now use the usual series for log(1+w).