Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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)) .....

$$\mathrm{les}\:\mathrm{points}\:\mathrm{singuliers}\:\mathrm{sont}\:\mathrm{les}\:\mathrm{poles}\:\mathrm{c}\:\mathrm{a}\:\mathrm{dire}\:\mathrm{1},\mathrm{e}^{\frac{\mathrm{2i}\pi}{\mathrm{3}}} \:\mathrm{et}\:\mathrm{e}^{−\frac{\mathrm{2i}\pi}{\mathrm{3}}} \:..... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com