find singular point for each (1)f(z)=(e^z /z^2 ) , (2)f(z)=((sinz)/z) , (3)f(z)=((1−cosz)/(sinz^2 )) , (4)f(z)=ln∣z∣ how can solve this help me sir


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Question Number 130416 by mohammad17 last updated on 25/Jan/21

$f i n d s i n g u l a r p o i n t f o r e a c h$ $(1) f (z) = \frac{e^{z}}{z^{2}}, (2) f (z) = \frac{s i n z}{z},$ $(3) f (z) = \frac{1 - c o s z}{{s i n z}^{2}}, (4) f (z) = l n ∣ z ∣$ $h o w c a n s o l v e t h i s h e l p m e s i r$
	Answered by mathmax by abdo last updated on 25/Jan/21
	$1) f(z)=(e^z /z^2 ) ,o is a singular point (alsodouble pole) and Res(f,o)=lim_(z→0) (1/((2−1)!)){z^2 f(z)}^((1)) =lim_(z→o) {e^z }^((1)) =lim_(z→0) e^z =1 another way we have f(z)=(1/z^2 )Σ_(n=0) ^∞ (z^n /(n!)) =Σ_(n=0) ^∞ (1/(n!))z^(n−2) =(1/z^2 )+(1/z) +(1/2) +(1/(3!))z +(1/(4!))z^2 +.... Res(f,o)=coefficient of((1/z))=1 2)f(z)=((sinz)/z) ,o is singular point also simple pole and Res(f,o)=lim_(z→0) zf(z)=lim_(z→0) sinz=0$
	$1) f (z) = \frac{e^{z}}{z^{2}}, o is a singular point (alsodouble pole)$ $and Res (f, o) = \lim_{z \to 0} \frac{1}{(2 - 1)!} {z^{2} f (z)}^{(1)} = \lim_{z \to o} {e^{z}}^{(1)}$ $= \lim_{z \to 0} e^{z} = 1$ $another way we have f (z) = \frac{1}{z^{2}} \sum_{n = 0}^{\infty} \frac{z^{n}}{n!}$ $= \sum_{n = 0}^{\infty} \frac{1}{n!} z^{n - 2} = \frac{1}{z^{2}} + \frac{1}{z} + \frac{1}{2} + \frac{1}{3!} z + \frac{1}{4!} z^{2} + . . . .$ $Res (f, o) = coefficient of (\frac{1}{z}) = 1$ $2) f (z) = \frac{sinz}{z}, o is singular point also simple pole and$ $Res (f, o) = \lim_{z \to 0} zf (z) = \lim_{z \to 0} sinz = 0$
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