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I need a comparing. I'd know the residue of

$$\frac{e^zāˆ’1}{z^2\sin(z)}; z_0=0$$

I have wrote down the series expansion of $e^z$ and $\sin(z)$ and then I've made in evidence $z^3...$ in the end I executed the limit operation for $z\to0$.

The result is $1/6$ Is It correct?

I would avoid the third derivation of function.

user91500
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GenKs
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2 Answers2

2

We have

$$\frac{e^z-1}{z^2\sin z}=\frac{z+z^2/2+\cdots}{z^2(z-z^3/6+\cdots)}=\frac{z+z^2/2+\cdots}{z^3}\times(1+z^2/6+\cdots)\\=\frac 1{z^2}+\frac {\frac 12} z+\cdots$$ hence the residue is $\frac 1 2$.

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Let$$(\ast)=\frac{\mathrm e^z-1}{z^2\sin z},$$ then $$(\ast)=\frac{z+z^2/2+z^3/6+O(z^4)}{z^3-z^5/6+O(z^7)}=\frac{1+z/2+z^2/6+O(z^3)}{z^2(1-z^2/6+O(z^4))},$$ hence $$(\ast)=\frac{1+z/2+z^2/3+O(z^3)}{z^2}=\frac1{z^2}+\frac1{2z}+\frac13+O(z).$$

Did
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  • Excuse me, I don't understand. What have you done in this step: $$\frac{1+z/2+z^2/6+O(z^3)}{z^2(1-z^2/6+O(z^4))}=\frac{1+z/2+z^2/3+O(z^3)}{z^2}$$ ā€“ GenKs Jan 14 '14 at 20:03
  • $1/(1-z^2/6)=1+z^2/6+O(z^4)$ then multiply by the numerator. ā€“ Did Jan 14 '14 at 20:12