i) Prove that $\tan h \space 2 y = \dfrac {2b}{1+a^2+b^2}$

ii) Separate into real and imaginary parts of $\log(3+4i)$

(i)

We have

$\tan(x+iy) = a+ib …..(1)$

Taking complex conjugate we get

$\tan(x-iy) = a-ib …..(2)\\ \tan i\space 2y = \tan[(x+iy) – (x-iy)] \\ =\dfrac {\tan⁡(x+iy) - \tan⁡(x-iy)}{1 + \tan⁡(x+iy).\tan⁡(x-iy)} $

Replacing values of $\tan(x+iy)$ and $\tan(x-iy) \\ ∴\tan ⁡i\space 2y=\dfrac {1+ib-(a-ib)}{1+(a+ib)(a-ib) }\\ ∴i \tan h\space 2y=\dfrac {1+ib-a+ib}{1+a^2+b^2 } \\ ∴i \tan h\space 2y= \dfrac {2ib}{1+a^2+b^2} \\ ∴\tan h\space2y= \dfrac {2b}{1+a^2+b^2}$

Hence proved.

(ii)

We know that

$\log x+iy = \log r+iθ $

Where $r=\sqrt{x^2+y^2}$ and $ θ=\tan^{-1}⁡ \dfrac yx $

Here $x=3$ and $y=4$

$∴r=\sqrt{3^2+4^2}=5 \\ θ=\tan^{-1} \dfrac 43 \\ ∴\log3+4i= \log5+i\tan^{-1} \dfrac 43 $

Where log 5 is real

and $i\tan^{-1} \dfrac 43$ is imaginary