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Question Number 148213 by mathmax by abdo last updated on 26/Jul/21
f(z)=((cosz)/(1−sin(z^2 ))) find residus of f
$$\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{cosz}}{\mathrm{1}−\mathrm{sin}\left(\mathrm{z}^{\mathrm{2}} \right)} \\ $$$$\mathrm{find}\:\mathrm{residus}\:\mathrm{of}\:\mathrm{f} \\ $$
Answered by puissant last updated on 26/Jul/21
sin(z^2 )∽z^2 −(z^6 /6) ⇒ −sin(z^2 )∽−z^2 +(z^6 /6) ⇒ 1−sin(z^2 )∽1−z^2 +(z^6 /6) ∽z^7 ((1/z^7 )−(1/z^5 )+(1/(6z))) ⇒ f(z)∽((cos(z))/(z^7 ((1/z^7 )−(1/z^5 )+(1/(6z))))) or cos(z)=Σ_(n=0) ^∞ (((−1)^n )/((2n)!))z^(2n) ∽1−(z^2 /(2!))+(z^4 /(4!))−(z^6 /(6!))+(z^8 /(8!))... ⇒f(z)∽(1/z^7 )(1−(z^2 /2)+(z^4 /(4!))−(z^6 /(6!))+(z^8 /(8!)))×(1/(((1/z^7 )−(1/z^5 )+(1/(6z))))) ⇒ f(z)∽((1/z^7 )−(1/(2!z^5 ))+(1/(4!z^3 ))−(1/(6!z)))×(1/(((1/z^7 )−(1/z^5 )+(1/(6z))))) Res(f)=−(1/(6!))×6=−(1/(120))...
$$\mathrm{sin}\left(\mathrm{z}^{\mathrm{2}} \right)\backsim\mathrm{z}^{\mathrm{2}} −\frac{\mathrm{z}^{\mathrm{6}} }{\mathrm{6}}\:\Rightarrow\:−\mathrm{sin}\left(\mathrm{z}^{\mathrm{2}} \right)\backsim−\mathrm{z}^{\mathrm{2}} +\frac{\mathrm{z}^{\mathrm{6}} }{\mathrm{6}} \\ $$$$\Rightarrow\:\mathrm{1}−\mathrm{sin}\left(\mathrm{z}^{\mathrm{2}} \right)\backsim\mathrm{1}−\mathrm{z}^{\mathrm{2}} +\frac{\mathrm{z}^{\mathrm{6}} }{\mathrm{6}}\:\backsim\mathrm{z}^{\mathrm{7}} \left(\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{7}} }−\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{5}} }+\frac{\mathrm{1}}{\mathrm{6z}}\right) \\ $$$$\Rightarrow\:\mathrm{f}\left(\mathrm{z}\right)\backsim\frac{\mathrm{cos}\left(\mathrm{z}\right)}{\mathrm{z}^{\mathrm{7}} \left(\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{7}} }−\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{5}} }+\frac{\mathrm{1}}{\mathrm{6z}}\right)} \\ $$$$\mathrm{or}\:\mathrm{cos}\left(\mathrm{z}\right)=\underset{\mathrm{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}\right)!}\mathrm{z}^{\mathrm{2n}} \\ $$$$\:\backsim\mathrm{1}−\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{2}!}+\frac{\mathrm{z}^{\mathrm{4}} }{\mathrm{4}!}−\frac{\mathrm{z}^{\mathrm{6}} }{\mathrm{6}!}+\frac{\mathrm{z}^{\mathrm{8}} }{\mathrm{8}!}… \\ $$$$\Rightarrow\mathrm{f}\left(\mathrm{z}\right)\backsim\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{7}} }\left(\mathrm{1}−\frac{\mathrm{z}^{\mathrm{2}} }{\mathrm{2}}+\frac{\mathrm{z}^{\mathrm{4}} }{\mathrm{4}!}−\frac{\mathrm{z}^{\mathrm{6}} }{\mathrm{6}!}+\frac{\mathrm{z}^{\mathrm{8}} }{\mathrm{8}!}\right)×\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{7}} }−\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{5}} }+\frac{\mathrm{1}}{\mathrm{6z}}\right)} \\ $$$$\Rightarrow\:\mathrm{f}\left(\mathrm{z}\right)\backsim\left(\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{7}} }−\frac{\mathrm{1}}{\mathrm{2}!\mathrm{z}^{\mathrm{5}} }+\frac{\mathrm{1}}{\mathrm{4}!\mathrm{z}^{\mathrm{3}} }−\frac{\mathrm{1}}{\mathrm{6}!\mathrm{z}}\right)×\frac{\mathrm{1}}{\left(\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{7}} }−\frac{\mathrm{1}}{\mathrm{z}^{\mathrm{5}} }+\frac{\mathrm{1}}{\mathrm{6z}}\right)} \\ $$$$\:\:\:\:\mathrm{Res}\left(\mathrm{f}\right)=−\frac{\mathrm{1}}{\mathrm{6}!}×\mathrm{6}=−\frac{\mathrm{1}}{\mathrm{120}}… \\ $$

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