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Question Number 130416 by mohammad17 last updated on 25/Jan/21
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
$${find}\:{singular}\:{point}\:{for}\:{each}\: \\ $$$$ \\ $$$$\left(\mathrm{1}\right){f}\left({z}\right)=\frac{{e}^{{z}} }{{z}^{\mathrm{2}} }\:\:\:\:\:\:\:,\:\:\:\:\left(\mathrm{2}\right){f}\left({z}\right)=\frac{{sinz}}{{z}}\:\:\:, \\ $$$$ \\ $$$$\left(\mathrm{3}\right){f}\left({z}\right)=\frac{\mathrm{1}−{cosz}}{{sinz}^{\mathrm{2}} }\:\:\:\:\:,\:\:\:\left(\mathrm{4}\right){f}\left({z}\right)={ln}\mid{z}\mid \\ $$$$ \\ $$$${how}\:{can}\:{solve}\:{this}\:{help}\:{me}\:{sir} \\ $$
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
$$\left.\mathrm{1}\right)\:\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{e}^{\mathrm{z}} }{\mathrm{z}^{\mathrm{2}} }\:\:\:,\mathrm{o}\:\mathrm{is}\:\mathrm{a}\:\mathrm{singular}\:\mathrm{point}\:\:\:\left(\mathrm{alsodouble}\:\mathrm{pole}\right) \\ $$$$\mathrm{and}\:\mathrm{Res}\left(\mathrm{f},\mathrm{o}\right)=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\mathrm{z}^{\mathrm{2}} \mathrm{f}\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{o}} \:\:\left\{\mathrm{e}^{\mathrm{z}} \right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{0}} \:\:\:\:\mathrm{e}^{\mathrm{z}} \:=\mathrm{1} \\ $$$$\mathrm{another}\:\mathrm{way}\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{2}} }\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{z}^{\mathrm{n}} }{\mathrm{n}!} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{n}!}\mathrm{z}^{\mathrm{n}−\mathrm{2}} \:=\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{z}}\:+\frac{\mathrm{1}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{3}!}\mathrm{z}\:+\frac{\mathrm{1}}{\mathrm{4}!}\mathrm{z}^{\mathrm{2}} +…. \\ $$$$\mathrm{Res}\left(\mathrm{f},\mathrm{o}\right)=\mathrm{coefficient}\:\mathrm{of}\left(\frac{\mathrm{1}}{\mathrm{z}}\right)=\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{sinz}}{\mathrm{z}}\:\:,\mathrm{o}\:\mathrm{is}\:\mathrm{singular}\:\mathrm{point}\:\mathrm{also}\:\mathrm{simple}\:\mathrm{pole}\:\mathrm{and} \\ $$$$\mathrm{Res}\left(\mathrm{f},\mathrm{o}\right)=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{0}} \mathrm{zf}\left(\mathrm{z}\right)=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{0}} \mathrm{sinz}=\mathrm{0} \\ $$

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