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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) = \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} + \dots .

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