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Question Number 193670 by mokys last updated on 17/Jun/23
show that ∫_c e^(1/z^2 ) dz = 0 when ∣z∣ <1
$${show}\:{that}\:\int_{{c}} \:{e}^{\frac{\mathrm{1}}{{z}^{\mathrm{2}} }} \:{dz}\:=\:\mathrm{0}\:{when}\:\mid{z}\mid\:<\mathrm{1} \\ $$
Commented by mokys last updated on 17/Jun/23
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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 ■
$${You}\:{need}\:{to}\:{be}\:{aware}\:{that}\:{for}\:\mid{z}\mid<\mathrm{1}\:\:{except}\:{for}\: \\ $$$${z}=\mathrm{0}\:,\:{the}\:{function}\:{f}\left({z}\right)={e}^{\frac{\mathrm{1}}{{z}^{\mathrm{2}} }} \:{is}\:{a}\:{power}\:{series}. \\ $$$${However}\:{dont}\:{try}\:{to}\:{use}\:{Residue}\:{theorem}!!!\: \\ $$$${as}\:{z}=\mathrm{0}\:{is}\:{an}\:{isolated}\:{essential}\:{singularity}. \\ $$$${i}.{e},\:{for}\:{z}\neq\mathrm{0}\:{and}\:\mid{z}\mid<\mathrm{1}\:,\:{e}^{\frac{\mathrm{1}}{{z}^{\mathrm{2}} }} =\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{z}^{\mathrm{2}{n}} } \\ $$$${If}\:{C}:\:\mid{z}\mid=\mathrm{1}, \\ $$$${I}=\oint_{{C}} {e}^{\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\:} {dz}=\oint_{{C}} \left(\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{z}^{\mathrm{2}{n}} }\right)=\mathrm{0}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left\{\oint_{{C}} \frac{\mathrm{1}}{{z}^{\mathrm{2}{n}} }{dz}\right\} \\ $$$${C}\::\:\mid{z}\mid=\mathrm{1}\:{in}\:{polar}\:{form}\:{looks}\:{like}, \\ $$$$\left({r}=\mathrm{1},\mathrm{0}\leqslant\theta\leqslant\mathrm{2}\pi\right)\:{so}\:{z}\:=\:{e}^{{i}\theta} \:,\:\:\mathrm{0}\leqslant\theta\leqslant\mathrm{2}\pi \\ $$$${I}=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left\{\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \frac{\mathrm{1}}{{e}^{\mathrm{2}{in}\theta} }\:.{i}.{e}^{{i}\theta} \:{d}\theta\right\}={i}.\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left\{\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \frac{{d}\theta}{{e}^{\left(\mathrm{2}{n}−\mathrm{1}\right){i}\theta} }\right\} \\ $$$$={i}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left\{\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} {e}^{\left(\mathrm{1}−\mathrm{2}{n}\right){i}\theta} {d}\theta\right\}=\left(\frac{\mathrm{1}}{{i}}.\frac{{i}}{\mathrm{1}−\mathrm{2}{n}}\right)\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left\{{e}^{\left(\mathrm{1}−\mathrm{2}{n}\right){i}\theta} \right\}_{\mathrm{0}} ^{\mathrm{2}\pi} \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−\mathrm{2}{n}}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left({e}^{\left(\mathrm{1}−\mathrm{2}{n}\right)\mathrm{2}\pi{i}} −\mathrm{1}\right)=\mathrm{0}\:\:\:\:\:\blacksquare \\ $$$$ \\ $$$$ \\ $$

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