A) \[\pi +2{{\tan }^{-1}}x\]
B) \[\pi -2{{\tan }^{-1}}x\]
C) \[-\pi +2{{\tan }^{-1}}x\]
D) \[-\pi -2{{\tan }^{-1}}x\]
Correct Answer: B
Solution :
Let \[z=\,i\log \left( \frac{x-i}{x+i} \right)\]\[\Rightarrow \,\frac{z}{i}=\log \left( \frac{x-i}{x+i} \right)\] \[\Rightarrow \frac{z}{i}=\log \,\left[ \frac{x-i}{x+i}\times \frac{x-i}{x-i} \right]\]\[=\,\log \,\left[ \frac{{{x}^{2}}-1-2ix}{{{x}^{2}}+1} \right]\] \[\Rightarrow \frac{z}{i}=\log \left[ \frac{{{x}^{2}}-1}{{{x}^{2}}+1}-i\frac{2x}{{{x}^{2}}+1} \right]\] ......(i) \[\because \,\log (a+ib)=\log (r{{e}^{i\theta }})=\log r+i\theta \] = \[\log \sqrt{{{a}^{2}}+{{b}^{2}}}+i{{\tan }^{-1}}(b/a)\] Hence, \[\frac{z}{i}=\log \sqrt{{{\left( \frac{{{x}^{2}}-1}{{{x}^{2}}+1} \right)}^{2}}+{{\left( \frac{-2x}{{{x}^{2}}+1} \right)}^{2}}}+i{{\tan }^{-1}}\left( \frac{-2x}{{{x}^{2}}-1} \right)\] [by eqn. (i)] \[\frac{z}{i}=\log \frac{\sqrt{{{x}^{4}}+1-2{{x}^{2}}+4{{x}^{2}}}}{{{({{x}^{2}}+1)}^{2}}}\] \[+i{{\tan }^{-1}}\left( \frac{2x}{1-{{x}^{2}}} \right)\] \[=\log 1+i\,(2{{\tan }^{-1}}x)\]\[=0+i\,(2{{\tan }^{-1}}x)\] \[\therefore z={{i}^{2}}2{{\tan }^{-1}}x=-2{{\tan }^{-1}}x\]\[=\pi -2{{\tan }^{-1}}x\].
You need to login to perform this action.
You will be redirected in
3 sec
