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Question Number 158005 by mkam last updated on 30/Oct/21

Commented by mkam last updated on 30/Oct/21

$$howcanitprovethishelpmeplease?$$

Commented by mkam last updated on 30/Oct/21

$$?????$$

Commented by benhamimed last updated on 30/Oct/21

$$z=1+\sin \theta +i\cos \theta$$$$=1+\cos (\frac{\pi }{2}-\theta )+i\sin (\frac{\pi }{2}-\theta )$$$$=1+e^{i(\frac{\pi }{2}-\theta )}$$$$=e^{i(\frac{\pi }{4}-\frac{\theta }{2})}(e^{-i(\frac{\pi }{4}-\frac{\theta }{2})}+e^{i(\frac{\pi }{4}-\frac{\theta }{2})})$$$$=2\cos (\frac{\pi }{4}-\frac{\theta }{2}).e^{i(\frac{\pi }{4}-\frac{\theta }{2})}$$$${\left[\frac{1+\sin \theta +i\cos \theta }{1+\sin \theta -i\cos \theta }\right]}^n={\left[\frac{z}{\stackrel{-}{z}}\right]}^n$$$$={\left[e^{i(\frac{\pi }{2}-\theta )}\right]}^n=e^{i(\frac{n\pi }{2}-n\theta )}$$$$=\cos (\frac{n\pi }{2}-n\theta )+i\sin (\frac{n\pi }{2}-n\theta )$$

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