Let $x=r \cos \theta$ .....................(1)

$y=r \sin \theta$ .............................(2).

Here $x=3, y=4.$ Squaring and adding (1) and (2), we get $x^2+y^2=r^2 \Rightarrow r= \sqrt{x^2+y^2}.$

Dividing (1) by (2), we get $\tan \theta= \dfrac{y}{x} \Rightarrow \theta=\tan^{-1} \bigg(\dfrac{y}x\bigg).$

Thus, 

$log(x+iy)=log[r(cos \theta+isin \theta)]$

 $=[log(r)+log(cos \theta+isin \theta)]$

$=log r+log[cos(2n \pi+\theta)+isin(2n \pi+\theta)]$

$=logr+log[e^{i(2n \pi+\theta)}]$    (using D'Moivre's Theorem)

$=logr+i(2n \pi+\theta) $  (Since, $log_e^{e^x}=x$ )

$=log \sqrt{(3^2+4^2)}+i(2n\pi+tan^{-1} {4 \over 3})$

$=log5+i(2n\pi+tan^{-1} {4 \over 3})$       Answer.