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I am a bit stuck on this problem:

Find all the singularities in the finite plane and the corresponding residues. Show the details.

$$\frac{8}{1+z^2}$$

So I know the residual is going to be the coefficient associated with the first negative exponent in the laurent series.

So the singularities are at $z = i $ because $i^2 = -i$ and that's when the denominator will equal 0.

But how do I find the residuals?

So I remember that: $$\frac{1}{1+z} = 1 - z + z^2 - ...$$

So I now multiply each element by 8:

$$\frac{8}{1+z} = 8(1 - z + z^2 - ...)t$$

But none of these terms have an exponent of -1 so I cannot find the coefficient?

Sigh, what am. I doing wrong?

Next problem, same as above but different function: $$\frac{1}{1-e^z}$$

Jwan622
  • 5,704

2 Answers2

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$$\frac{8}{Z^2+1} = \frac{8}{(Z+i)(Z-i)} = \frac{A}{Z+i}+\frac{B}{Z-i}$$

$$=\frac{A(Z-i) + B(Z+i)}{Z^2+1}$$

So $$(A+B)Z + (B-A)i = 8$$

So $$A+B=0$$ $$(B-A)i=8$$

So $$B=-4i$$ $$A=4i$$

So, $$\frac{8}{Z^2+1}=\frac{4i}{Z+i} + \frac{-4i}{Z-i}$$

Math Keeps Me Busy
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If f has a simple zero at a, then, near a, $f(z) = (z-a)f'(a) + (z-a)^2 \frac{f''(a)}{2}+... = (z-a)f'(a) (1 + O(z-a)) $ ,

so:

$\frac{1}{f(z)} = \frac{1}{(z-a)f'(a)(1 + O(z-a))} = \frac{1}{(z-a)f'(a)} \frac{1}{(1 + O(z-a))} = \frac{1}{(z-a)f'(a)} (1 + O(z-a)) = \frac{1}{(z-a)f'(a)} + O(1) $

from where you can read the residue of 1/f at a.

VictorZurkowski
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  • Sorry mind relating this to my problem? – Jwan622 Nov 30 '20 at 14:50
  • Just a bit, but here it is: $\frac{8}{1+z^2} = \frac{1}{f(z)}$ with $f(z) = (1+z^2)/8$, which has simple zeros at $a = \pm i$. Take the simple zero $a=i$, there $f'(i)= 2z/8\Big|_{z=i} = i/4$, so, near $a$, $\frac{8}{1+z^2} = \frac{1}{(z-i)f'(i)} + $[higher powers of $(z-i)$ ] = $\frac{4}{(z-i)i} + ...$, hence Res{$\frac{8}{1+z^2}$ at $a=i$} = $\frac{4}{i} = -4i$. – VictorZurkowski Nov 30 '20 at 21:21
  • In the case of $\frac{1}{1-e^z}$ we have $f(z) = 1-e^z$, which has simple zeroes at $a=2\pi k\ i$ with $k \in \mathbb{Z}$. Take one of the zeros, since $f'(2\pi k\ i) = e^z\Big|_{z=2\pi k\ i} = 1$, we have $\frac{1}{1-e^z} = \frac{1}{(z-2\pi k\ i)f'(2\pi k\ i)} + ... = \frac{1}{(z-2\pi k\ i)} + ... $, so Res{$\frac{1}{1-e^z}$ at $a=2\pi k\ i$} = $1$ (which is the coefficient of $(z-2\pi k\ i)^{-1}$ in the Laurant expansion of the function under consideration in a neighborhood of $a=2\pi k\ i$) – VictorZurkowski Nov 30 '20 at 21:36