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Question Number 86550 by DuDono last updated on 29/Mar/20

$$\mathrm{show}\mathrm{that}$$$$\tan {}^{-1}x=\frac{\ln (-x^2+2ix+1)-\ln (x^2+1)}{2i}$$

Commented by john santu last updated on 29/Mar/20

$$\mathrm{y}=\tan \mathrm{x}=\frac{\sin \mathrm{x}}{\cos \mathrm{x}}$$$$\mathrm{y}=\frac{{\mathrm{e}}^{\mathrm{ix}}-{\mathrm{e}}^{-\mathrm{ix}}}{2\mathrm{i}}\times \frac{2\mathrm{i}}{{\mathrm{e}}^{\mathrm{ix}}+{\mathrm{e}}^{-\mathrm{ix}}}$$$$\mathrm{y}=\frac{{\mathrm{e}}^{\mathrm{ix}}-{\mathrm{e}}^{-\mathrm{ix}}}{{\mathrm{e}}^{\mathrm{ix}}+{\mathrm{e}}^{-\mathrm{ix}}}\Rightarrow {\mathrm{ye}}^{\mathrm{ix}}+{\mathrm{ye}}^{-\mathrm{ix}}={\mathrm{e}}^{\mathrm{ix}}-{\mathrm{e}}^{-\mathrm{ix}}$$$$(\mathrm{y}+1){\mathrm{e}}^{-\mathrm{ix}}={\mathrm{e}}^{\mathrm{ix}}(1-\mathrm{y})[{\mathrm{e}}^{\mathrm{ix}}=\mathrm{w}]$$$$\frac{\mathrm{y}+1}{\mathrm{w}}=\mathrm{w}(1-\mathrm{y})\Rightarrow {\mathrm{w}}^2(1-\mathrm{y})-(1+\mathrm{y})=0$$$$\mathrm{w}=\pm \sqrt{\frac{\mathrm{y}+1}{\mathrm{y}-1}}$$$$\mathrm{you}\mathrm{can}\mathrm{solve}\mathrm{it}$$

Commented by Prithwish Sen 1 last updated on 29/Mar/20

$$\mathrm{RHS}$$$$\frac{\ln \frac{{(1+\mathrm{ix})}^2}{\{1-{\left(\mathrm{ix}\right)}^2\}}}{2\mathrm{i}}=\frac{\ln (1+\mathrm{ix})-\ln (1-\mathrm{ix})}{2\mathrm{i}}$$$$=\frac{\{\ln \sqrt{1+{\mathrm{x}}^2}+\mathrm{i}(2\mathrm{n}\pi +{\tan }^{-1}\mathrm{x})\}-\{\ln \sqrt{1+{\mathrm{x}}^2}+\mathrm{i}(2\mathrm{n}\pi -{\tan }^{-1}\mathrm{x})\}}{2\mathrm{i}}$$$$={\mathbf{tan}}^{-1}\mathbf{x}\mathbf{please}\mathbf{check}.$$

Commented by mathmax by abdo last updated on 30/Mar/20

$$wehaveprovedthatarctanx=\frac{1}{2i}ln\left(\frac{1+ix}{1-ix}\right)forx\in C$$$$lettakexreal\Rightarrow arctan\left(x\right)=\frac{1}{2i}ln\left(\frac{{(1+ix)}^2}{1+x^2}\right)$$$$=\frac{1}{2i}\{ln(1+2ix-x^2)-ln(1+x^2)\}$$$$=\frac{ln(-x^2+2ix+1)-ln(1+x^2)}{2i}sotheresultisproved.$$

Answered by Ar Brandon last updated on 30/Mar/20

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