Question Number 130392 by mathmax by abdo last updated on 25/Jan/21
$$\mathrm{find}\:\int_{\mid\mathrm{z}\mid=\mathrm{1}} \:\:\frac{\mathrm{1}−\mathrm{cosz}}{\mathrm{z}^{\mathrm{2}} }\mathrm{dz} \\ $$
Answered by mohammad17 last updated on 25/Jan/21
$${hello}\:{sir}\:{can}\:{you}\:{help}\:{me}\:{in}\:{Q}\:\mathrm{130416} \\ $$
Answered by mathmax by abdo last updated on 25/Jan/21
![∫_(∣z∣=1) ((1−cosz)/z^2 )dz =2iπ Res(f,0) with f(z)=((1−cosz)/z^2 ) o is double pole ⇒Res(f,o)=lim_(z→0) (1/((2−1)!)){z^2 f(z)}^((1)) =lim_(z→0) (1−cosz)^((1)) =lim_(z→0) sinz =0 another way ∣z∣=1 ⇒z=e^(iθ) ⇒∫_(∣z∣=1) ((1−cosz)/z^2 ) =∫_0 ^(2π) ((1−cos(e^(iθ) ))/e^(2iθ) )ie^(iθ) dθ =i∫_0 ^(2π) e^(−iθ) (1−cos(e^(iθ) ))dθ =i∫_0 ^(2π) e^(−iθ) dθ−i∫_0 ^(2π) e^(−iθ) cos(e^(iθ) )dθ =0−i∫_0 ^(2π) e^(−iθ) (Σ_(n=0) ^∞ (((−1)^n e^(2inθ) )/((2n)!)))dθ =−iΣ_(n=0) ^∞ (((−1)^n )/(2n!)) ∫_0 ^(2π) e^((2n−1)iθ) dθ =−iΣ_(n=0) ^∞ (((−1)^n )/((2n)!))[(1/((2n−1)))e^((2n−1)iθ) ]_0 ^(2π) =0](https://www.tinkutara.com/question/Q130464.png)
$$\int_{\mid\mathrm{z}\mid=\mathrm{1}} \:\:\frac{\mathrm{1}−\mathrm{cosz}}{\mathrm{z}^{\mathrm{2}} }\mathrm{dz}\:=\mathrm{2i}\pi\:\mathrm{Res}\left(\mathrm{f},\mathrm{0}\right)\:\mathrm{with}\:\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{1}−\mathrm{cosz}}{\mathrm{z}^{\mathrm{2}} } \\ $$$$\mathrm{o}\:\mathrm{is}\:\mathrm{double}\:\mathrm{pole}\:\Rightarrow\mathrm{Res}\left(\mathrm{f},\mathrm{o}\right)=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{0}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\mathrm{z}^{\mathrm{2}} \mathrm{f}\left(\mathrm{z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{0}} \:\:\:\left(\mathrm{1}−\mathrm{cosz}\right)^{\left(\mathrm{1}\right)} \:=\mathrm{lim}_{\mathrm{z}\rightarrow\mathrm{0}} \:\mathrm{sinz}\:=\mathrm{0} \\ $$$$\mathrm{another}\:\mathrm{way}\:\mid\mathrm{z}\mid=\mathrm{1}\:\Rightarrow\mathrm{z}=\mathrm{e}^{\mathrm{i}\theta} \:\Rightarrow\int_{\mid\mathrm{z}\mid=\mathrm{1}} \:\:\frac{\mathrm{1}−\mathrm{cosz}}{\mathrm{z}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{e}^{\mathrm{i}\theta} \right)}{\mathrm{e}^{\mathrm{2i}\theta} }\mathrm{ie}^{\mathrm{i}\theta} \mathrm{d}\theta \\ $$$$=\mathrm{i}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \mathrm{e}^{−\mathrm{i}\theta} \left(\mathrm{1}−\mathrm{cos}\left(\mathrm{e}^{\mathrm{i}\theta} \right)\right)\mathrm{d}\theta\:=\mathrm{i}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\mathrm{e}^{−\mathrm{i}\theta} \:\mathrm{d}\theta−\mathrm{i}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\mathrm{e}^{−\mathrm{i}\theta} \:\mathrm{cos}\left(\mathrm{e}^{\mathrm{i}\theta} \right)\mathrm{d}\theta \\ $$$$=\mathrm{0}−\mathrm{i}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\mathrm{e}^{−\mathrm{i}\theta} \left(\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{\mathrm{2in}\theta} }{\left(\mathrm{2n}\right)!}\right)\mathrm{d}\theta \\ $$$$=−\mathrm{i}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{2n}!}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\mathrm{e}^{\left(\mathrm{2n}−\mathrm{1}\right)\mathrm{i}\theta} \:\:\mathrm{d}\theta \\ $$$$=−\mathrm{i}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}\right)!}\left[\frac{\mathrm{1}}{\left(\mathrm{2n}−\mathrm{1}\right)}\mathrm{e}^{\left(\mathrm{2n}−\mathrm{1}\right)\mathrm{i}\theta} \right]_{\mathrm{0}} ^{\mathrm{2}\pi} \:=\mathrm{0} \\ $$