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How can I find the residue of the function below at $z_0=-2i$

$$f(z) = (z^2+4iz)\sin(\frac{3}{z+2i})$$

I tried expressing $f(z)$ in terms of taylor series.

$$f(z) = (z^2+4iz).\sum_{k=0}^{\infty} \left(\frac{\sin(-2i)(-1)^k}{(2k)!}((\frac{3}{z+2i})+2i)^{2k}+\frac{\cos(-2i)(-1)^{k+1}}{(2k+1)!}((\frac{3}{z+2i})+2i)^{2k+1}\right)$$

I know we are supposed to find out $b_1$, which is the coefficient of $\frac{1}{z+2i}$ but it seems that it is infinite sum. Can someone please advise?

Thanks in advance!

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Hint. We have that $$f(z)=(4+(z+2i)^2)\sum_{k=0}^{\infty}\frac{(-1)^k}{(2k+1)!}\left(\frac{3}{z+2i}\right)^{2k+1}\\ =(4+(z+2i)^2)\cdot\left(\frac{3}{z+2i}-\frac{1}{6}\frac{27}{(z+2i)^3}+O\left(\frac{1}{(z+2i)^5}\right)\right).$$ Are you able to find the coefficient of $(z+2i)^{-1}$ in the above product?

P.S. The final result should be $\frac{15}{2}$.

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  • $\begingroup$ Why is it that you can write your sine function as such a simple expression? Isn't the taylor series of sine function the complicated expression as I have written above? $\endgroup$ – javabeginner_0101 Oct 26 '16 at 8:57
  • $\begingroup$ @javabeginner_0101 In this case it is convenient to use the fact that $\sin w=\sum_{k=0}^{\infty}(-1)^k w^{2k+1}/(2k+1)!$ for all complex numbers $w$. Of course you can expand the sine also at other points. $\endgroup$ – Robert Z Oct 26 '16 at 9:00
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You are overcomplicating things. Write $$\sin\left(\frac{3}{z+2i}\right)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!}\left(\frac{3}{z+2i}\right)^{2n+1}=\sum_{n=0}^\infty \frac{(-1)^n\cdot3^{2n+1}}{(2n+1)!}\left(z+2i\right)^{-2n-1},$$ So that it is already in the Laurent expansion at $z_0=-2i$. Now rewrite the first polynomial as a polynomial in $(z+2i)$, i.e write $$z^2+4iz=(z+2i)^2+4.$$ Now multiply these then read the coefficient $a_{-1}$.

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  • $\begingroup$ @boon Why is it that you can write your sine function that way? Isn't the taylor series of sine the complicated big expression like the one I written? $\endgroup$ – javabeginner_0101 Oct 26 '16 at 8:56

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