After working out the poles to:

$\displaystyle

f(z)=\frac{1}{2z^{2}+(6-4i)z-12i}$

I have the answer z1=2i and z2=3

Given this how would we find the taylor series if:

$\displaystyle \left | z \right |< \left | z1 \right |$

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- May 4th 2010, 09:19 AMzizou1089Taylor Series expansion
After working out the poles to:

$\displaystyle

f(z)=\frac{1}{2z^{2}+(6-4i)z-12i}$

I have the answer z1=2i and z2=3

Given this how would we find the taylor series if:

$\displaystyle \left | z \right |< \left | z1 \right |$ - May 4th 2010, 09:50 AMFailure
Shouldn't we have $\displaystyle z_2=-3$, or have you given us the wrong $\displaystyle f(z)$? For the rest of this post I am assuming that your $\displaystyle f(z)$ is right, and that $\displaystyle z_2=-3$.

Quote:

Given this how would we find the taylor series if:

$\displaystyle \left | z \right |< \left | z1 \right |$

$\displaystyle f(z)=\frac{a}{2i-z}+\frac{b}{3+z}=-\frac{1}{2(2i-z)(3+z)}$

which gives, if I am not misaken, that $\displaystyle a=b=-\frac{1}{6+4i}$. (I might be mistaken, however, so please to the math yourself.)

Now you can develop each of these two fractions with the help of a geometric series like this:

$\displaystyle \frac{a}{2i-z}+\frac{b}{3+z} = \frac{a}{2i}\cdot \frac{1}{1-\frac{z}{2i}}+\frac{b}{3}\cdot \frac{1}{1-\left(-\frac{z}{3}\right)}$

$\displaystyle =\frac{a}{2i}\sum_{n=0}^\infty \left(\frac{z}{2i}\right)^n+\frac{b}{3}\sum_{n=0}^ \infty \left(-\frac{z}{3}\right)^n=\ldots$ - May 4th 2010, 12:46 PMzizou1089
Ok, wouldnt $\displaystyle \left |2i \right |$=2?

- May 4th 2010, 01:14 PMzizou1089
How would the answer change if $\displaystyle \left |z1 \right |< \left | z \right |< \left | z2 \right |$

- May 4th 2010, 08:06 PMFailure
In that case, you need a Laurent-Series development, because the geometric series that we used for the first term of the partial fraction decomposition no longer converges for such values of z.

You can get it, starting from the same partial fraction decomposition, like this

$\displaystyle \frac{a}{2i-z}+\frac{b}{3+z} = -\frac{a}{z}\cdot \frac{1}{1-\frac{2i}{z}}+\frac{b}{3}\cdot \frac{1}{1-\left(-\frac{z}{3}\right)}$

$\displaystyle =-\frac{a}{z}\sum_{n=0}^\infty \left(\frac{2i}{z}\right)^n+\frac{b}{3}\sum_{n=0}^ \infty \left(-\frac{z}{3}\right)^n=\ldots $ - May 5th 2010, 03:08 AMzizou1089
Okay Im starting to get the hang of this, its just aint $\displaystyle \left | 2i \right |=2$? For the first part when we divide $\displaystyle \left | z \right |< \left | z1 \right |$ by $\displaystyle \left | z1 \right | $ we have

$\displaystyle \frac{\left | z \right |}{\left | 2i \right |}< 1$

which is equal to $\displaystyle \frac{\left | z \right |}{2}< 1$

therefore:

$\displaystyle \frac{a}{2}\cdot \frac{1}{1-\frac{z}{2}}+\frac{b}{3}\cdot \frac{1}{1-\left(-\frac{z}{3}\right)}$?? - May 5th 2010, 05:58 AMFailure
Great!

Quote:

its just aint $\displaystyle \left | 2i \right |=2$? For the first part when we divide $\displaystyle \left | z \right |< \left | z1 \right |$ by $\displaystyle \left | z1 \right | $ we have

$\displaystyle \frac{\left | z \right |}{\left | 2i \right |}< 1$

which is equal to $\displaystyle \frac{\left | z \right |}{2}< 1$

therefore:

$\displaystyle \frac{a}{2}\cdot \frac{1}{1-\frac{z}{2}}+\frac{b}{3}\cdot \frac{1}{1-\left(-\frac{z}{3}\right)}$??

Therefore, if $\displaystyle |z| < |2i|$, we have that $\displaystyle \left|\frac{z}{2i}\right| < 1$, and thus the series $\displaystyle \sum_{n=0}^\infty \left(\frac{z}{2i}\right)^n$ converges. But if $\displaystyle |2i|< |z|$, we have $\displaystyle 1<\left|\frac{z}{2i}\right|$, thus $\displaystyle \sum_{n=0}^\infty \left(\frac{z}{2i}\right)^n$ diverges.

By a purely algebraic transformation of that first term of the partial fraction decomposition we can use the (for $\displaystyle |2i| < |z|$) convergent geometric series $\displaystyle \sum_{n=0}^\infty \left(\frac{2i}{z}\right)^n$, instead.

Simililarly, if we were told to develop the given function for $\displaystyle |-3|<|z|$, we would have to transform the second term of the partial fraction decomposition in such a way that we can develop it in a, for $\displaystyle \left|\frac{-3}{z}\right|<1$, convergent series of the form $\displaystyle \sum_{n=0}^\infty\left(-\frac{3}{z}\right)^n$. - May 5th 2010, 08:37 AMzizou1089
- May 5th 2010, 09:18 AMFailure
In this case you cannot develop the series around $\displaystyle 0$ anymore, but have instead to develop it around $\displaystyle z_1$: which means, that you want to develop your function into Laurent-series that do not have terms of the form $\displaystyle a_n z^n$ but terms of the form $\displaystyle a_n (z-z_1)^n$ instead.

The first term of your partial fraction decomposition is already (almost) in that form, namely

$\displaystyle \frac{a}{2i-z}=-a(z-2i)^{-1}$.

For the second term of your partial fraction decomposition you can substitute $\displaystyle (z-2i)+2i$ for $\displaystyle z$ and try to transform it into a power series in terms of $\displaystyle (z-2i)^n$. - May 6th 2010, 05:34 PMzizou1089
- May 6th 2010, 07:58 PMFailure
I have only very little time for this (it's early in the morning here in Switzerland). So let me start with the partial fraction decomposition that we've already used successfully for the other cases and try to develop it as series of integral powers of $\displaystyle (z-z_1)$, that is $\displaystyle (z-2i)$:

$\displaystyle \frac{a}{2i-z}+\frac{b}{3+z} = -a(z-2i)^{-1}+\frac{b}{3+2i+(z-2i)} =-a(z-2i)^{-1}+\frac{b}{3+2i}\cdot\frac{1}{1-\left(-\frac{z-2i}{3+2i}\right)}$

$\displaystyle =-a(z-2i)^{-1}+\frac{b}{3+2i}\cdot\sum_{n=0}^\infty \left(-\frac{z-2i}{3+2i}\right)^n=\ldots$

Note that this series converges, as required, if $\displaystyle |z-2i| < |-3-2i|=|3+2i|$, because in that case we have that $\displaystyle \left|-\frac{z-2i}{3+2i}\right|<1$ - May 7th 2010, 05:18 AMzizou1089