The question is 2 parts - I'm to find the Laurent series valid for $$1 < |z| <4$$ and $$|z| > 5$$ I've already solved the first part, but I have a conceptual question about the second part.

My result for the first part:

I found that after putting the $f(z)$ into the standard geometric series form, it resulted in

$$-\sum_{n=0}^\infty \frac{z^n}{2^{2n+1}}; |z|<4$$ $$-\sum_{n=0}^\infty \frac{2^{2n-1}}{z^n}; |z|>4$$ $$-\sum_{n=0}^\infty (-3)^nz^n; |z|<1$$ $$-\sum_{n=0}^\infty \frac{(-3)^n}{z^n}; |z|>1$$


$$f(z)=-\sum_{n=0}^\infty \frac{z^n}{2^{2n+1}}-\sum_{n=0}^\infty \frac{(-3)^n}{z^n} $$ for $1<|z|<4$.

The way I understand this is that a Laurent series is simply another way to represent a function, $f(z)$, but in a way that puts into focus what happens around the singularities of the function.

My attempt for part 2:

Given my answer above and my reasoning for what it means, I'd say that this sum is the answer. It explains what happens at all radius greater than $4$.

$$f(z)=-\sum_{n=0}^\infty \frac{2^{2n-1}}{z^n}; |z|>4$$

Is this the correct way to understand this?

  • $\begingroup$ Why do you believe that the $\frac 3{z+1}$ term has no contribution for $|z| > 4$? $\endgroup$ – Paul Sinclair Feb 27 '16 at 5:47
  • $\begingroup$ @PaulSinclair Oh I see what you're saying. So for $|z|>5$ it would be the sum of $|z|>1$ and $|z|>4$, right? I originally thought $|z|>4$ alone would cover it. $\endgroup$ – whatwhatwhat Mar 2 '16 at 2:19
  • 1
    $\begingroup$ What I am saying is that $f$ is the sum of $\frac 2{z-4}$ and $\frac{-3}{z+1}$. You sum functions, not domains, which are sets. In each of the three domains $\{z\ :\ |z| <1\}, \{z\ :\ 1 < |z| <4\}, \{z\ :\ 4 < |z| \}$ (there is no reason at all to restrict that last domain to $|z| > 5$), both functions have a Laurent series. As $f$ is their sum, the Laurent series of $f$ is the sum of the Laurent series of these two functions. Therefore, you calculate the Laurent series of the constituent functions in each domain, then add them together, as I did in my answer. $\endgroup$ – Paul Sinclair Mar 2 '16 at 3:27

For $|z| < 4$,

$$\begin{align}\frac 2{z - 4} &= \frac 2{-4}\frac 1{1 - \frac z 4}\\ &=\frac {-1}2\sum_{n=0}^\infty \left(\frac z4\right)^n\\ &=-\sum_{n=0}^\infty\frac {z^n}{2^{2n+1}}\end{align}$$

For $|z| > 4$, $$\begin{align}\frac 2{z - 4} &= \frac 2{z}\frac 1{1 - \frac 4 z}\\ &=\frac 2z\sum_{n=0}^\infty \left(\frac 4z\right)^n\\ &=\sum_{n=0}^\infty\frac {2^{2n+1}}{z^{n+1}}\\ &= \sum_{n=1}^\infty\frac {2^{2n-1}}{z^{n}}\end{align}$$

for $|z| < 1$,$$\begin{align}\frac {-3}{z + 1} &= -3\frac 1{1 - (-z)}\\ &=-3\sum_{n=0}^\infty (-z)^n\\ &=-3\sum_{n=0}^\infty(-1)^nz^n\end{align}$$ for $|z| > 1$ $$\begin{align}\frac {-3}{z + 1} &= \frac {-3}z\frac 1{1 - (-\frac 1z)}\\ &=\frac {-3}z\sum_{n=0}^\infty \left(-\frac 1z\right)^n\\ &=3\sum_{n=0}^\infty\frac {(-1)^{n+1}}{z^{n+1}}\\ &=3\sum_{n=1}^\infty\frac {(-1)^{n}}{z^{n}}\end{align}$$

Hence for $|z| < 1$, $$f(z) = -\sum_{n=0}^\infty\frac {z^n}{2^{2n+1}} - 3\sum_{n=0}^\infty(-1)^nz^n = -\sum_{n=0}^\infty \frac{1 + 3(-1)^n}{2^{2n+1}}z^n$$ For $1 < |z| < 4$, $$f(z) = -\sum_{n=0}^\infty\frac {z^n}{2^{2n+1}} + \sum_{n=1}^\infty\frac {3(-1)^{n}}{z^{n}}$$ For $4 < |z|$, $$f(z) = \sum_{n=1}^\infty\frac {2^{2n-1}}{z^{n}} + 3\sum_{n=1}^\infty\frac {(-1)^{n}}{z^{n}} = \sum_{n=1}^\infty \frac{2^{2n-1} + 3(-1)^n}{z^n}$$


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