show-that-c-e-1-z-2-dz-0-when-z-lt-1-

Question Number 193670 by mokys last updated on 17/Jun/23

s h o w t h a t \int_{c} e^{\frac{1}{z^{2}}} d z = 0 w h e n ∣ z ∣ < 1

Commented by mokys last updated on 17/Jun/23

? ? ? ? ?

Answered by Rajpurohith last updated on 19/Jun/23

Y o u n e e d t o b e a w a r e t h a t f o r ∣ z ∣< 1 e x c e p t f o r

z = 0, t h e f u n c t i o n f (z) = e^{\frac{1}{z^{2}}} i s a p o w e r s e r i e s .

H o w e v e r d o n t t r y t o u s e R e s i d u e t h e o r e m!!!

a s z = 0 i s a n i s o l a t e d e s s e n t i a l s i n g u l a r i t y .

i . e, f o r z \neq 0 a n d ∣ z ∣< 1, e^{\frac{1}{z^{2}}} = 1 + \underset{n = 1}{\sum^{\infty}} \frac{1}{z^{2 n}}

I f C : ∣ z ∣= 1,

I = \oint_{C} e^{\frac{1}{z^{2}}} d z = \oint_{C} (1 + \underset{n = 1}{\sum^{\infty}} \frac{1}{z^{2 n}}) = 0 + \underset{n = 1}{\sum^{\infty}} {\oint_{C} \frac{1}{z^{2 n}} d z}

C : ∣ z ∣= 1 i n p o l a r f o r m l o o k s l i k e,

(r = 1, 0 ⩽ θ ⩽ 2 π) s o z = e^{i θ}, 0 ⩽ θ ⩽ 2 π

I = \underset{n = 0}{\sum^{\infty}} {\int_{0}^{2 π} \frac{1}{e^{2 i n θ}} . i . e^{i θ} d θ} = i . \underset{n = 0}{\sum^{\infty}} {\int_{0}^{2 π} \frac{d θ}{e^{(2 n - 1) i θ}}}

= i \underset{n = 0}{\sum^{\infty}} {\int_{0}^{2 π} e^{(1 - 2 n) i θ} d θ} = (\frac{1}{i} . \frac{i}{1 - 2 n}) \underset{n = 0}{\sum^{\infty}} {e^{(1 - 2 n) i θ}}_{0}^{2 π}

= \frac{1}{1 - 2 n} \underset{n = 0}{\sum^{\infty}} (e^{(1 - 2 n) 2 π i} - 1) = 0