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Question Number 97576 by Rio Michael last updated on 08/Jun/20

$$\mathrm{Show}\mathrm{that}RE\left[\frac{1}{1-z}\right]=\frac{1}{2}\mathrm{where}z=\cos \theta +i\sin \theta$$$$$$

Answered by smridha last updated on 08/Jun/20

$$\mathit{R}\mathit{E}\left[\frac{1}{1-{\mathit{e}}^{\mathit{i}\mathit{\theta }}}\right]=\mathit{R}\mathit{E}\left[\frac{1+e^{i\mathit{\theta }}}{1-e^{2i\mathit{\theta }}}\right]$$$$\mathit{R}E\left[\frac{e^{-i\mathit{\theta }}+1}{e^{-i\mathit{\theta }}-e^{i\mathit{\theta }}}\right]=Re\left[\frac{(cos\mathit{\theta }-isin\theta +1)}{-2\mathit{i}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }}\right]$$$$=\mathit{R}\mathit{E}\left[\frac{sin\mathit{\theta }+i(\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }+1)}{2\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }}\right]=\mathit{R}\mathit{E}[\frac{1}{2}+i\left(\frac{\mathit{c}\mathit{o}\mathit{s}\mathit{\theta }+1}{2sin\theta }\right)]$$$$=\frac{1}{2}$$

Commented by Rio Michael last updated on 08/Jun/20

$$\mathrm{really}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{understand}\mathrm{this}\mathrm{sir}...\mathrm{can}\mathrm{you}\mathrm{please}$$$$\mathrm{take}\mathrm{it}\mathrm{step}\mathrm{by}\mathrm{step}?$$

Commented by smridha last updated on 08/Jun/20

$$\mathit{f}\mathit{r}\mathit{o}\mathit{m}\mathit{w}\mathit{h}\mathit{i}\mathit{c}\mathit{h}\mathit{s}\mathit{t}\mathit{e}\mathit{p}\mathit{y}\mathit{o}\mathit{u}\mathit{g}\mathit{e}\mathit{t}\mathit{p}\mathit{u}\mathit{z}\mathit{z}\mathit{l}\mathit{e}\mathit{d}?$$$$\mathit{I}\mathit{t}\mathit{h}\mathit{i}\mathit{n}\mathit{k}\mathit{y}\mathit{o}\mathit{u}\mathit{t}\mathit{a}\mathit{k}\mathit{e}\mathit{s}\mathit{o}\mathit{m}\mathit{e}\mathit{t}\mathit{i}\mathit{m}\mathit{e}$$$$\mathit{a}\mathit{n}\mathit{d}\mathit{t}\mathit{r}\mathit{y}\mathit{t}\mathit{o}\mathit{u}\mathit{n}\mathit{d}\mathit{e}\mathit{r}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{d}\mathit{s}\mathit{l}\mathit{o}\mathit{w}\mathit{l}\mathit{y}$$$$\mathit{s}\mathit{l}\mathit{o}\mathit{w}\mathit{l}\mathit{y}.....\mathit{t}\mathit{h}\mathit{i}\mathit{s}\mathit{i}\mathit{s}\mathit{t}\mathit{h}\mathit{e}\mathit{t}\mathit{r}\mathit{i}\mathit{c}\mathit{k}..$$

Commented by Rio Michael last updated on 08/Jun/20

$$\mathbf{am}\mathbf{fine}\mathbf{now}\mathbf{sir}\mathbf{thanks}.$$

Answered by abdomathmax last updated on 08/Jun/20

$$\frac{1}{1-\mathrm{z}}=\frac{1}{1-\cos \theta -\mathrm{isin}\theta }=\frac{1-\cos \theta +\mathrm{isin}\theta }{{(1-\cos \theta )}^2+{\sin }^2\theta }$$$$=\frac{1-\cos \theta +\mathrm{isin}\theta }{1-2\cos \theta +1}=\frac{1-\cos \theta }{2(1-\cos \theta )}+\mathrm{i}\frac{\sin \theta }{2(1-\cos \theta )}$$$$=\frac{1}{2}+\mathrm{i}\frac{2\sin \left(\frac{\theta }{2}\right)\cos \left(\frac{\theta }{2}\right)}{2\times {2\sin }^2\left(\frac{\theta }{2}\right)}=\frac{1}{2}+\frac{\mathrm{i}}{2}\mathrm{cotan}\left(\frac{\theta }{2}\right)$$$$\Rightarrow \mathrm{Re}\left(\frac{1}{1-\mathrm{z}}\right)=\frac{1}{2}\mathrm{and}\mathrm{Im}\left(\frac{1}{1-\mathrm{z}}\right)=\frac{1}{2}\mathrm{cotan}\left(\frac{\theta }{2}\right)$$

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