show that ∫_c e^(1/z^2 ) dz = 0 when ∣z∣ <1


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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
	$You need to be aware that for ∣z∣<1 except for z=0 , the function f(z)=e^(1/z^2 ) is a power series. However dont try to use Residue theorem!!! as z=0 is an isolated essential singularity. i.e, for z≠0 and ∣z∣<1 , e^(1/z^2 ) =1+Σ_(n=1) ^∞ (1/z^(2n) ) If C: ∣z∣=1, I=∮_C e^((1/z^2 ) ) dz=∮_C (1+Σ_(n=1) ^∞ (1/z^(2n) ))=0+Σ_(n=1) ^∞ {∮_C (1/z^(2n) )dz} C : ∣z∣=1 in polar form looks like, (r=1,0≤θ≤2π) so z = e^(iθ) , 0≤θ≤2π I=Σ_(n=0) ^∞ {∫_0 ^( 2π) (1/e^(2inθ) ) .i.e^(iθ) dθ}=i.Σ_(n=0) ^∞ {∫_0 ^( 2π) (dθ/e^((2n−1)iθ) )} =iΣ_(n=0) ^∞ {∫_0 ^( 2π) e^((1−2n)iθ) dθ}=((1/i).(i/(1−2n)))Σ_(n=0) ^∞ {e^((1−2n)iθ) }_0 ^(2π) =(1/(1−2n)) Σ_(n=0) ^∞ (e^((1−2n)2πi) −1)=0 ■$
	$Y o u n e e d t o b e a w a r e t h a t f o r ∣ z ∣< 1 e x c e p t f o r$ $z = 0, t h e f u n c t i o n f (z) = e^{\frac{1}{z^{2}}} i s a p o w e r s e r i e s .$ $H o w e v e r d o n t t r y t o u s e R e s i d u e t h e o r e m!!!$ $a s z = 0 i s a n i s o l a t e d e s s e n t i a l s i n g u l a r i t y .$ $i . e, f o r z \neq 0 a n d ∣ z ∣< 1, e^{\frac{1}{z^{2}}} = 1 + \underset{n = 1}{\sum^{\infty}} \frac{1}{z^{2 n}}$ $I f C : ∣ z ∣= 1,$ $I = \oint_{C} e^{\frac{1}{z^{2}}} d z = \oint_{C} (1 + \underset{n = 1}{\sum^{\infty}} \frac{1}{z^{2 n}}) = 0 + \underset{n = 1}{\sum^{\infty}} {\oint_{C} \frac{1}{z^{2 n}} d z}$ $C : ∣ z ∣= 1 i n p o l a r f o r m l o o k s l i k e,$ $(r = 1, 0 ⩽ θ ⩽ 2 π) s o z = e^{i θ}, 0 ⩽ θ ⩽ 2 π$ $I = \underset{n = 0}{\sum^{\infty}} {\int_{0}^{2 π} \frac{1}{e^{2 i n θ}} . i . e^{i θ} d θ} = i . \underset{n = 0}{\sum^{\infty}} {\int_{0}^{2 π} \frac{d θ}{e^{(2 n - 1) i θ}}}$ $= i \underset{n = 0}{\sum^{\infty}} {\int_{0}^{2 π} e^{(1 - 2 n) i θ} d θ} = (\frac{1}{i} . \frac{i}{1 - 2 n}) \underset{n = 0}{\sum^{\infty}} {e^{(1 - 2 n) i θ}}_{0}^{2 π}$ $= \frac{1}{1 - 2 n} \underset{n = 0}{\sum^{\infty}} (e^{(1 - 2 n) 2 π i} - 1) = 0 ◼$
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