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Question Number 145946 by tabata last updated on 09/Jul/21
the type of singular point of f(z)=((cos(πz))/((1−z^3 ))) is ?
$${the}\:{type}\:{of}\:{singular}\:{point}\:{of}\:{f}\left({z}\right)=\frac{{cos}\left(\pi{z}\right)}{\left(\mathrm{1}−{z}^{\mathrm{3}} \right)}\:{is}\:? \\ $$$$ \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 09/Jul/21
les points singuliers de f sont les residus def z^3 −1=0 ⇔(z−1)(z^2 +z+1)=0 ⇔z=1 or z^2 +z+1=0 z^2 +z+1=0→Δ=−3 ⇒z_1 =((−1+i(√3))/2)=e^((2iπ)/3) z_2 =((−1−i(√3))/2)=e^(−((2iπ)/3)) ⇒f(z)=((cos(πz))/((1−z)(z−z_1 )(z−z_2 ))) Res(f,1) =((cos(π))/3)=−(1/3) Res(f,z_1 )=((cos(πz_1 ))/((1−z_1 )(z_1 −z_2 )))=((cos(πe^((2iπ)/3) ))/((1−e^((i2π)/3) )(i(√3)))) Res(f,z_2 )=((cos(πz_2 ))/((1−z_2 )(−i(√3)))) rest de terminer le calcul....
$$\mathrm{les}\:\mathrm{points}\:\mathrm{singuliers}\:\mathrm{de}\:\mathrm{f}\:\mathrm{sont}\:\mathrm{les}\:\mathrm{residus}\:\mathrm{def} \\ $$$$\mathrm{z}^{\mathrm{3}} −\mathrm{1}=\mathrm{0}\:\Leftrightarrow\left(\mathrm{z}−\mathrm{1}\right)\left(\mathrm{z}^{\mathrm{2}} +\mathrm{z}+\mathrm{1}\right)=\mathrm{0}\:\Leftrightarrow\mathrm{z}=\mathrm{1}\:\mathrm{or}\:\mathrm{z}^{\mathrm{2}} \:+\mathrm{z}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{z}^{\mathrm{2}} \:+\mathrm{z}+\mathrm{1}=\mathrm{0}\rightarrow\Delta=−\mathrm{3}\:\Rightarrow\mathrm{z}_{\mathrm{1}} =\frac{−\mathrm{1}+\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \\ $$$$\mathrm{z}_{\mathrm{2}} =\frac{−\mathrm{1}−\mathrm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \:\Rightarrow\mathrm{f}\left(\mathrm{z}\right)=\frac{\mathrm{cos}\left(\pi\mathrm{z}\right)}{\left(\mathrm{1}−\mathrm{z}\right)\left(\mathrm{z}−\mathrm{z}_{\mathrm{1}} \right)\left(\mathrm{z}−\mathrm{z}_{\mathrm{2}} \right)} \\ $$$$\mathrm{Res}\left(\mathrm{f},\mathrm{1}\right)\:=\frac{\mathrm{cos}\left(\pi\right)}{\mathrm{3}}=−\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$\mathrm{Res}\left(\mathrm{f},\mathrm{z}_{\mathrm{1}} \right)=\frac{\mathrm{cos}\left(\pi\mathrm{z}_{\mathrm{1}} \right)}{\left(\mathrm{1}−\mathrm{z}_{\mathrm{1}} \right)\left(\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} \right)}=\frac{\mathrm{cos}\left(\pi\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \right)}{\left(\mathrm{1}−\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \right)\left(\mathrm{i}\sqrt{\mathrm{3}}\right)} \\ $$$$\mathrm{Res}\left(\mathrm{f},\mathrm{z}_{\mathrm{2}} \right)=\frac{\mathrm{cos}\left(\pi\mathrm{z}_{\mathrm{2}} \right)}{\left(\mathrm{1}−\mathrm{z}_{\mathrm{2}} \right)\left(−\mathrm{i}\sqrt{\mathrm{3}}\right)} \\ $$$$\mathrm{rest}\:\mathrm{de}\:\mathrm{terminer}\:\mathrm{le}\:\mathrm{calcul}…. \\ $$
Commented by mathmax by abdo last updated on 09/Jul/21
les points singuliers sont les poles c a dire 1,e^((2iπ)/3) et e^(−((2iπ)/3)) .....
$$\mathrm{les}\:\mathrm{points}\:\mathrm{singuliers}\:\mathrm{sont}\:\mathrm{les}\:\mathrm{poles}\:\mathrm{c}\:\mathrm{a}\:\mathrm{dire}\:\mathrm{1},\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \:\mathrm{et}\:\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \:….. \\ $$

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