We have

$\tan \left ( \dfrac {\pi}{4}+iv \right ) = re^{i\theta} = r (\cos \theta + i \sin \theta)$

$ \therefore \dfrac {\tan \left ( \frac {\pi}{4} \right )\tan (iv)}{1-\tan \left ( \frac {\pi}{4} \right )\tan (iv)} = r (\cos \theta + i \sin \theta)$

$ \therefore \dfrac {1+\tan (iv)}{1-\tan (iv)} = r (\cos \theta + i \sin \theta)$

$ \therefore \dfrac {1+i \tan hv}{1-i\tan hv} = r (\cos \theta + i \sin \theta) \ \cdots (A)$

$\therefore \dfrac {(1+i\tan hv)}{(1-i \tan hv)} \cdot \dfrac {(1+i\tanh v)}{(1+i \tanh v)} = r (\cos \theta + i \sin \theta) $

$ \therefore \dfrac {(1-\tanh^2 v + 2i \tan hv)}{(1+\tanh^2 v)}=r(\cos \theta + i \sin \theta) $

$\therefore \dfrac {(\text{sech}^2 v + 2i \tan hv)}{(1+\tanh^2 v)} = r (\cos \theta + i \sin \theta)$

On comparing real and imaginary parts,

$r \cos \theta = \dfrac {(\text{sech}^2v)}{(1+\tanh^2 v)} \ \cdots (1)$

$ r \sin \theta = \dfrac {(2\tan hv)}{(1+\tanh^2 v)} \ \cdots (2)$

Squarign (1) and (2) and adding,

$\therefore r^2 (\cos ^2 \theta + \sin ^2 \theta ) = \dfrac {(\text{sech}^4v)+4\tanh^2 v}{(1+\tanh^2v)^2}$

$\therefore r^2 = \dfrac {(1-\tanh^2 v)^2 + 4\tanh^2 v}{(1+\tanh^2 v)^2} = \dfrac {(1+\tanh^2 v)^2}{(1+\tanh^2v)}$

Hence, r=1

(ii)

Dividing (2) by (1)

$\tan \theta = \dfrac {2 \tanh v}{\text{sech}^2 v} = \dfrac {2 \tanh v \cosh^2 v}{1} = \sinh2v $

Hence proved, $\tan \theta = \sinh 2v$

(iii)

Using (A) and putting r=1

We get,

$\therefore \dfrac {1+i \tanh v}{1-i \tanh v}=e^{i\theta}$

Applying componendo and divinendo, we get,

$\therefore \dfrac {2}{2 i \tanh v} = \dfrac {e^{i\theta}+1}{e^{i\theta}-1}$

$= \dfrac {e^{\frac {i\theta}{2}}+e^{-\frac {i\theta}{2}}}{e^{\frac {i\theta}{2}}- e^{-\frac {i\theta}{2}}}$

$ = \dfrac {2 \cos \left ( \frac {\theta}{2} \right )}{ 2 i\sin \left ( \frac {\theta}{2} \right )}$

Hence, $\tanh v = \tan \left ( \dfrac {\theta }{2} \right ) $