Let-p-and-q-be-positive-integers-having-no-positive-common-divisors-except-unity-Let-z-1-z-2-z-q-be-the-q-values-of-z-p-q-where-z-is-a-fixed-complex-number-Then-the-product-z-1-z-2-z

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Question Number 140248 by EnterUsername last updated on 05/May/21

$$\mathrm{Let}p\mathrm{and}q\mathrm{be}\mathrm{positive}\mathrm{integers}\mathrm{having}\mathrm{no}\mathrm{positive}$$$$\mathrm{common}\mathrm{divisors}\mathrm{except}\mathrm{unity}.\mathrm{Let}{z}_1,z_2,\dots ,z_q\mathrm{be}\mathrm{the}$$$$q\mathrm{values}\mathrm{of}{z}^{p/q},\mathrm{where}z\mathrm{is}\mathrm{a}\mathrm{fixed}\mathrm{complex}\mathrm{number}.\mathrm{Then}$$$$\mathrm{the}\mathrm{product}{z}_1{z}_2\dots z_q\mathrm{is}\mathrm{equal}\mathrm{to}$$$$\left(\mathrm{A}\right)z^p,\mathrm{if}q\mathrm{is}\mathrm{odd}\left(\mathrm{B}\right)-z^p,\mathrm{if}q\mathrm{is}\mathrm{even}$$$$\left(\mathrm{C}\right)z^p,\mathrm{if}q\mathrm{is}\mathrm{even}\left(\mathrm{D}\right)-z^p,\mathrm{if}q\mathrm{is}\mathrm{odd}$$

Answered by mr W last updated on 05/May/21

$$z={re}^{\theta i}={re}^{(2k\pi +\theta )i}$$$$z_k=z^{p/q}=r^{p/q}{e}^{\frac{(2k\pi +\theta )p}{q}i},k=0,1,\dots ,q-1$$$$\mathrm{\Pi }{z}_k=r^p{e}^{\frac{(2\pi \times \frac{q(q-1)}{2}+q\theta )p}{q}i}=r^p{e}^{(q\pi -\pi +\theta )pi}$$$$={\left({re}^{(q\pi -\pi +\theta )i}\right)}^p$$$$={\left({re}^{\theta i}{e}^{(q-1)\pi i}\right)}^p$$$$={\left({ze}^{(q-1)\pi i}\right)}^p$$$$=z^p{e}^{p(q-1)\pi i}$$$$=\{\begin{array}{l}{z}^{p}ifp(q-1)iseven\\ -z^{p}ifp(q-1)isodd\end{array}$$$$$$$$\left(A\right)ifqisold,p(q-1)iseven,$$$$resultis{z}^p\Rightarrow correct$$$$$$$$\left(B\right)ifqiseven,thenpisodd,then$$$$p(q-1)isodd,theresultis-z^p,$$$$\Rightarrow correct.$$

Commented by EnterUsername last updated on 05/May/21

$$\mathrm{Thank}\mathrm{you},\mathrm{Sir}$$

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