For fz-sinzz2- the residue of the pole z-0 is

I: Given f(z)=sinzz^2=1z^2 [z z^33!+z^55!.... ] ⇒ f(z) = 1z z3!+z^35!.... ∴ residue of the pole (at z=0) = Coefficient of ( 1z 0 ) in power series expans ...

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Engineering Mathematics

Residue Formula

For fz=sinzz2...

Question

For $f (z) = \frac{s i n (z)}{z^{2}}$ , the residue of the pole $z = 0$ is

1

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Solution

The correct option is A 1

I:
Given
$f (z) = \frac{s i n z}{z^{2}} = \frac{1}{z^{2}} [z - \frac{z^{3}}{3!} + \frac{z^{5}}{5!} . . . .]$
$\Rightarrow$ $f (z) = \frac{1}{z} - \frac{z}{3!} + \frac{z^{3}}{5!} . . . .$
$∴$ residue of the pole (at $z = 0$ ) = Coefficient of $(\frac{1}{z - 0})$ in power series expansion $= 1$

II:
$f (z) = \frac{s i n z}{z^{2}} = \frac{1}{z^{2}} [z - \frac{z^{3}}{3!} + \frac{z^{5}}{5!} . . . .]$
$\Rightarrow$ $f (z) = \frac{1}{z} - \frac{z}{3!} + \frac{z^{3}}{5!} - . . . .$
So $Z = 0$ is pole of order $1$ (smile pole)
So, $[R e s f (z); (z = 0)] = {lim}_{z \to 0} (Z - 0) f (z)$
$= {lim}_{z \to 0} (\frac{s i n Z}{Z}) = 1$

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For f(z)=sin(z)z2, the residue of the pole z=0 is 1

For $f (z) = \frac{s i n (z)}{z^{2}}$ , the residue of the pole $z = 0$ is

1