(Complex Analysis) Residue of essential singularity

[polar flint]

Residue of sin(1/z)(5-z+3z^2-4z^3) at 0?

[glad zenith]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with: ```diff
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[junior shale]

To find the residue of the function $f(z) = sin(1/z)(5 - z + 3z² - 4z³) $ at $z = 0$, we need to find the coefficient of the $z^{-1}$ term in the Laurent series expansion of $f(z)$ around $z = 0$.

First, we can write the function $f(z)$ as:

$f(z) = sin(1/z)(5 - z + 3z² - 4z³)$

Now we can expand $\sin\left(\frac{1}{z}\right)$ as a power series:

$sin(1/z) = (1/z) - (1/3!z³) + (1/5!z⁵) - ...$

Substituting this series into the expression for $f(z)$,we get:

$f(z) = (5/z² - 1/z + 3 - 4z) - (1/3!)(5/z⁴ - 1/z³ + 3/z² - 4/z) + ...$

The coefficient of $z^{-1}$ in this expansion is given by the coefficient of $1/z$ in the term $\frac{1}{3!}\left(\frac{5}{z^4} - \frac{1}{z^3} + \frac{3}{z^2} - \frac{4}{z}\right)$. Therefore, the residue of $f(z)$ at $z=0$ is:

$Res(f,0) = -1/3!*(-4) = 2/3$

Therefore, the residue of $\sin\left(\frac{1}{z}\right)(5-z+3z^2-4z^3)$ at $z=0$ is $\frac{2}{3}$.

[ruby forge]

7eventy6ix

[junior shale]

Hope this helps @polar flint

[polar flint]

Dont give me no chatgpt thanks

[junior shale]

Answers right there.

But ok.

[polar flint]

I cant be sure its right because chatgpt is wrong for math

[junior shale]

Fair enough, I see where your coming from. You need to close the chat to get another member to help you. And re open a submission.

+close