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Question Number 124595 by physicstutes last updated on 04/Dec/20

$$\mathrm{Given}\mathrm{that}\omega =e^{i\theta },\theta \ne n\pi ,n\in \mathbb{N}$$$$\mathrm{show}\mathrm{that}$$$$\left(1\right)\frac{{\omega }^2-1}{\omega }=2i\sin \theta$$$$\left(2\right){(1+\omega )}^n=2^n{\cos }^n\left(\frac{1}{2}\theta \right)e^{\frac{1}{2}\left(in\theta \right)}$$

Answered by Ar Brandon last updated on 04/Dec/20

$$\mathrm{f}\left(\omega \right)=\frac{{\omega }^2-1}{\omega }=\frac{{\mathrm{e}}^{2\mathrm{i}\theta }-1}{{\mathrm{e}}^{\mathrm{i}\theta }}={\mathrm{e}}^{\mathrm{i}\theta }-{\mathrm{e}}^{-\mathrm{i}\theta }$$$$=(\cos \theta +\mathrm{isin}\theta )-(\cos (-\theta )+\mathrm{isin}(-\theta ))$$$$=(\cos \theta +\mathrm{isin}\theta )-(\cos \theta -\mathrm{isin}\theta )$$$$=2\mathrm{isin}\theta$$

Commented by physicstutes last updated on 04/Dec/20

$$\mathrm{thats}\mathrm{great}$$

Answered by mnjuly1970 last updated on 04/Dec/20

$$1:{\omega }^2-1=(cos\left(2\theta \right)+isin\left(2\theta \right))-1=2isin\left(\theta \right)cos\left(\theta \right)-2{sin}^2\left(\theta \right)$$$$=2isin\left(\theta \right)[cos\left(\theta \right)+isin\left(\theta \right)]$$$$\Rightarrow \frac{{\omega }^2-1}{\omega }=2isin\left(\theta \right)$$$$corallary:\mathrm{\Omega }={\int }_0^{\frac{\pi }{2}}ln\left(Im\left(\frac{{\omega }^2-1}{2w}\right)\right)d\theta =-\frac{\pi }{2}ln\left(2\right)✓$$

Answered by Ar Brandon last updated on 04/Dec/20

$$\mathrm{f}\left(\omega \right)={(1+\omega )}^{\mathrm{n}}={(1+\cos \theta +\mathrm{isin}\theta )}^{\mathrm{n}}$$$$={({2\cos }^2\frac{\theta }{2}+2\mathrm{isin}\frac{\theta }{2}\cos \frac{\theta }{2})}^{\mathrm{n}}$$$$={\left[2\cos \frac{\theta }{2}(\cos \frac{\theta }{2}+\mathrm{isin}\frac{\theta }{2})\right]}^{\mathrm{n}}$$$$=2^{\mathrm{n}}{\cos }^{\mathrm{n}}\frac{\theta }{2}{\mathrm{e}}^{\mathrm{i}\frac{\mathrm{n}\theta }{2}}$$

Commented by Ar Brandon last updated on 04/Dec/20

$$1={\cos }^2\frac{\theta }{2}+{\sin }^2\frac{\theta }{2}$$$$\cos \theta ={\cos }^2\frac{\theta }{2}-{\sin }^2\frac{\theta }{2}$$$$\sin \theta =2\sin \frac{\theta }{2}\cos \frac{\theta }{2}$$$$\cos \frac{\theta }{2}+\mathrm{isin}\frac{\theta }{2}={\mathrm{e}}^{\mathrm{i}\frac{\theta }{2}}$$

Answered by mnjuly1970 last updated on 04/Dec/20

$$2.{(1+\omega )}^n={(1+cos\left(\theta \right)+isin\left(\theta \right))}^n$$$$={(2{cos}^2\left(\frac{\theta }{2}\right)+2isin\left(\frac{\theta }{2}\right)cos\left(\frac{\theta }{2}\right))}^n$$$$=2^n\left({cos}^n\left(\frac{\theta }{2}\right)\right){(cos\left(\frac{\theta }{2}\right)+isin\left(\frac{\theta }{2}\right))}^n$$$$=2^n{cos}^n\left(\frac{\theta }{2}\right)e^{i\left(\frac{n\theta }{2}\right)}✓$$

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