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Question Number 196303 by sonukgindia last updated on 22/Aug/23

Answered by sniper237 last updated on 22/Aug/23

$$\underset{n=1}{\sum ^{\mathrm{\infty }}}Im{\left(\frac{e^{i\pi /4}}{2}\right)}^n=Im\left(\underset{n=1}{\sum ^{\mathrm{\infty }}}{\left(e^{i\pi /4}/2\right)}^n\right)$$$$=Im\left(\frac{e^{i\pi /4}/2}{1-e^{i\pi /4}/2}\right)$$

Commented by sonukgindia last updated on 22/Aug/23

$$explainthis$$

Commented by JDamian last updated on 22/Aug/23

$$1.\sin \alpha =\mathrm{Im}\left\{{\mathrm{e}}^{i\alpha }\right\}$$$$2.\mathrm{\Sigma }\left(\mathrm{Im}\left\{z_i\right\}\right)=\mathrm{Im}\left\{\mathrm{\Sigma }{z}_i\right\}$$

Answered by Mathspace last updated on 22/Aug/23

$$s\left(x\right)=\sum _{n=1}^{\mathrm{\infty }}\frac{sin\left(nx\right)}{2^n}\Rightarrow$$$$s\left(x\right)=Im\left(\sum _{n=1}^{\mathrm{\infty }}\frac{e^{inx}}{2^n}\right)but$$$$\sum _{n=1}^{\mathrm{\infty }}\frac{e^{inx}}{2^n}=\sum _{n=1}^{\mathrm{\infty }}{\left(\frac{1}{2}{e}^{ix}\right)}^n$$$$\frac{1}{1-\frac{1}{2}{e}^{ix}}(lookthat\mid \frac{1}{2}{e}^{ix}\mid <1)$$$$=\frac{2}{2-cosx-isinx}$$$$=\frac{2(2-cosx+isinx)}{{(2-cosx)}^2+{sin}^{2}x}$$$$\Rightarrow s\left(x\right)=\frac{2sinx}{4-4cosx+1}$$$$=\frac{2sinx}{5-4cosx}$$$$and\sum _{n=1}^{\mathrm{\infty }}\frac{sin\left(\frac{n\pi }{4}\right)}{2^n}$$$$=s\left(\frac{\pi }{4}\right)=\frac{2sin\left(\frac{\pi }{4}\right)}{5-4cos\left(\frac{\pi }{4}\right)}$$$$=\frac{2\times \frac{1}{\sqrt{2}}}{5-4.\frac{\sqrt{2}}{2}}=\frac{\sqrt{2}}{5-2\sqrt{2}}$$

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