A) 0
B) \[\infty \]
C) 1
D) None of these
Correct Answer: B
Solution :
\[\tan (u+iv)=i\]\[\Rightarrow \] \[\frac{\tan u+\tan (iv)}{1-\tan u\,\tan (iv)}=i\] \[\Rightarrow \]\[\tan u+i\tan \,hv=i[1-i\tan u\tan \,hv]\] \[[\because \,\tan ix=i\tan \,hx]\] \[\Rightarrow \] \[\tan u(1-\tan \,hv)=i(1-\tan \,hv)\] \[\Rightarrow \]\[(\tan u-i)\,(1-\tanh v)=0\] \[\Rightarrow \] \[1-\tan \,hv=0\]\[\Rightarrow \] \[\tan \,hv=1\] \[\Rightarrow \] \[\frac{{{e}^{v}}-{{e}^{-v}}}{{{e}^{v}}+{{e}^{-v}}}=1\]\[\Rightarrow \] \[{{e}^{v}}-{{e}^{-v}}={{e}^{v}}+{{e}^{-v}}\] \[\Rightarrow \] \[2{{e}^{-v}}=0\] \[\Rightarrow \,\,{{e}^{-v}}={{e}^{-\infty }}\] \[\Rightarrow \] \[v=\infty \].
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