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Question Number 145946 by tabata last updated on 09/Jul/21

$$thetypeofsingularpointoff\left(z\right)=\frac{cos\left(\pi z\right)}{(1-z^3)}is?$$$$$$$$$$

Answered by mathmax by abdo last updated on 09/Jul/21

$$\mathrm{les}\mathrm{points}\mathrm{singuliers}\mathrm{de}\mathrm{f}\mathrm{sont}\mathrm{les}\mathrm{residus}\mathrm{def}$$$${\mathrm{z}}^3-1=0\iff (\mathrm{z}-1)({\mathrm{z}}^2+\mathrm{z}+1)=0\iff \mathrm{z}=1\mathrm{or}{\mathrm{z}}^2+\mathrm{z}+1=0$$$${\mathrm{z}}^2+\mathrm{z}+1=0\to \mathrm{\Delta }=-3\Rightarrow {\mathrm{z}}_1=\frac{-1+\mathrm{i}\sqrt{3}}{2}={\mathrm{e}}^{\frac{2\mathrm{i}\pi }{3}}$$$${\mathrm{z}}_2=\frac{-1-\mathrm{i}\sqrt{3}}{2}={\mathrm{e}}^{-\frac{2\mathrm{i}\pi }{3}}\Rightarrow \mathrm{f}\left(\mathrm{z}\right)=\frac{\cos \left(\pi \mathrm{z}\right)}{(1-\mathrm{z})(\mathrm{z}-{\mathrm{z}}_1)(\mathrm{z}-{\mathrm{z}}_2)}$$$$\mathrm{Res}(\mathrm{f},1)=\frac{\cos \left(\pi \right)}{3}=-\frac{1}{3}$$$$\mathrm{Res}(\mathrm{f},{\mathrm{z}}_1)=\frac{\cos \left(\pi {\mathrm{z}}_1\right)}{(1-{\mathrm{z}}_1)({\mathrm{z}}_1-{\mathrm{z}}_2)}=\frac{\cos \left(\pi {\mathrm{e}}^{\frac{2\mathrm{i}\pi }{3}}\right)}{(1-{\mathrm{e}}^{\frac{\mathrm{i}2\pi }{3}})\left(\mathrm{i}\sqrt{3}\right)}$$$$\mathrm{Res}(\mathrm{f},{\mathrm{z}}_2)=\frac{\cos \left(\pi {\mathrm{z}}_2\right)}{(1-{\mathrm{z}}_2)(-\mathrm{i}\sqrt{3})}$$$$\mathrm{rest}\mathrm{de}\mathrm{terminer}\mathrm{le}\mathrm{calcul}....$$

Commented by mathmax by abdo last updated on 09/Jul/21

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

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