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Question Number 100943 by Ar Brandon last updated on 29/Jun/20

Determine the poles of the function; f(x)=((x^5 −1)/(x^3 −1))

$$\mathcal{D}\mathrm{etermine}\:\mathrm{the}\:\mathrm{poles}\:\mathrm{of}\:\mathrm{the}\:\mathrm{function}; \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}^{\mathrm{5}} −\mathrm{1}}{\mathrm{x}^{\mathrm{3}} −\mathrm{1}} \\ $$

Answered by mathmax by abdo last updated on 29/Jun/20

the poles are z with verify z^3 −1 =0 and z≠1 z^3 =1 ⇔z^3 =e^(i(2kπ)) ⇒z_k =e^(i((2kπ)/3)) and k ∈{1,2} ⇒the poles are z_1 =e^((i2π)/3) and z_2 =e^((i4π)/3)

$$\mathrm{the}\:\mathrm{poles}\:\mathrm{are}\:\mathrm{z}\:\:\mathrm{with}\:\mathrm{verify}\:\mathrm{z}^{\mathrm{3}} −\mathrm{1}\:=\mathrm{0}\:\:\mathrm{and}\:\mathrm{z}\neq\mathrm{1} \\ $$$$\mathrm{z}^{\mathrm{3}} \:=\mathrm{1}\:\Leftrightarrow\mathrm{z}^{\mathrm{3}} \:=\mathrm{e}^{\mathrm{i}\left(\mathrm{2k}\pi\right)} \:\Rightarrow\mathrm{z}_{\mathrm{k}} =\mathrm{e}^{\mathrm{i}\frac{\mathrm{2k}\pi}{\mathrm{3}}} \:\:\:\mathrm{and}\:\mathrm{k}\:\in\left\{\mathrm{1},\mathrm{2}\right\}\:\Rightarrow\mathrm{the}\:\mathrm{poles}\:\mathrm{are}\: \\ $$$$\mathrm{z}_{\mathrm{1}} =\mathrm{e}^{\frac{\mathrm{i2}\pi}{\mathrm{3}}} \:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} =\mathrm{e}^{\frac{\mathrm{i4}\pi}{\mathrm{3}}} \\ $$

Commented by Ar Brandon last updated on 29/Jun/20

Thanks. I didn't understand the question.

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