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Question Number 66421 by Rio Michael last updated on 14/Aug/19

$$showthatforagivencomplexnumberz$$$$z^n=r^n(cosn\theta +isinn\theta )$$

Answered by MJS last updated on 14/Aug/19

$$z=r(\cos \theta +\mathrm{i}\sin \theta )=r{\mathrm{e}}^{\mathrm{i}\theta }$$$$z^n={\left(r{\mathrm{e}}^{\mathrm{i}\theta }\right)}^n=r^n{\mathrm{e}}^{\mathrm{i}n\theta }=r^n(\cos n\theta +\mathrm{i}\sin n\theta )$$

Commented by Rio Michael last updated on 15/Aug/19

$$thankssir$$$$$$

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