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Question Number 75178 by Rio Michael last updated on 08/Dec/19

givn that z = 1−i(√3) express z in the form    z = r(cosθ + isinθ), hence express  z^7  in the form re^(iθ)

$${givn}\:{that}\:{z}\:=\:\mathrm{1}−{i}\sqrt{\mathrm{3}}\:{express}\:{z}\:{in}\:{the}\:{form}\: \\ $$$$\:{z}\:=\:{r}\left({cos}\theta\:+\:{isin}\theta\right),\:{hence}\:{express} \\ $$$${z}^{\mathrm{7}} \:{in}\:{the}\:{form}\:{re}^{{i}\theta} \\ $$

Answered by mr W last updated on 08/Dec/19

z=1−(√3)i=2((1/2)−((√3)/2)i)=2(cos ((5π)/3)+i sin ((5π)/3))  =2e^((5πi)/3)   z^7 =2^7 e^((35πi)/3) =128e^((5πi)/3)

$${z}=\mathrm{1}−\sqrt{\mathrm{3}}{i}=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{i}\right)=\mathrm{2}\left(\mathrm{cos}\:\frac{\mathrm{5}\pi}{\mathrm{3}}+{i}\:\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{3}}\right) \\ $$$$=\mathrm{2}{e}^{\frac{\mathrm{5}\pi{i}}{\mathrm{3}}} \\ $$$${z}^{\mathrm{7}} =\mathrm{2}^{\mathrm{7}} {e}^{\frac{\mathrm{35}\pi{i}}{\mathrm{3}}} =\mathrm{128}{e}^{\frac{\mathrm{5}\pi{i}}{\mathrm{3}}} \\ $$

Commented by Rio Michael last updated on 08/Dec/19

thanks sir, i appreciate

$${thanks}\:{sir},\:{i}\:{appreciate} \\ $$

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