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Question Number 66421 by Rio Michael last updated on 14/Aug/19
show that for a given complex number z   z^n  = r^n  (cosnθ + isinnθ)
$${show}\:{that}\:{for}\:{a}\:{given}\:{complex}\:{number}\:{z} \\ $$$$\:{z}^{{n}} \:=\:{r}^{{n}} \:\left({cosn}\theta\:+\:{isinn}\theta\right)\: \\ $$
Answered by MJS last updated on 14/Aug/19
z=r(cos θ +i sin θ)=re^(iθ)   z^n =(re^(iθ) )^n =r^n e^(inθ) =r^n (cos nθ +i sin nθ)
$${z}={r}\left(\mathrm{cos}\:\theta\:+\mathrm{i}\:\mathrm{sin}\:\theta\right)={r}\mathrm{e}^{\mathrm{i}\theta} \\ $$$${z}^{{n}} =\left({r}\mathrm{e}^{\mathrm{i}\theta} \right)^{{n}} ={r}^{{n}} \mathrm{e}^{\mathrm{i}{n}\theta} ={r}^{{n}} \left(\mathrm{cos}\:{n}\theta\:+\mathrm{i}\:\mathrm{sin}\:{n}\theta\right) \\ $$
Commented by Rio Michael last updated on 15/Aug/19
thanks sir
$${thanks}\:{sir} \\ $$$$ \\ $$

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