Question

solve:

$log(a+ib)=$ (where $a>0,b>0$)

• A. $log\sqrt{a^2+b^2}$
• B. $\frac{1}{2}log(a^2+b^2)+ita{n}^{-1}\left(\frac{b}{a}\right)$
• C. $\frac{1}{2}log(a^2+b^2)-ita{n}^{-1}\left(\frac{b}{a}\right)$
• D. $\frac{1}{2}log(a^2+b^2)+ita{n}^{-1}\left(\frac{a}{b}\right)$

Answer 1

The correct option is B $\frac{1}{2}log(a^2+b^2)+ita{n}^{-1}\left(\frac{b}{a}\right)$

Log $(a+ib)=\omega =x+iy$. say
$\therefore$ By laws of logarithm
$e^{x+iy}=a+ib$
$\therefore e^x.e^{iy}=a+ib$
$\therefore e^x[cis.y]=a+ib$
$\Rightarrow \left(e^{x}cosy\right)+\left(e^{x}siny\right)i=a+ib$
equating the real & imaginary part,
$a=e^{x}cosy$------(1)
$b=e^{x}siny$------(2)
To find ${}^{\prime }{x}^{\prime }$, square (1) & (2) & add them
$\therefore e^{2x}(si{n}^{2}y+co{s}^{2}y)=a^2+b^2$
$\therefore e^{2x}=a^2+b^2$
take log, $log{e}^{2x}=log(a^2+b^2)$
$\therefore 2xloge=log(a^2+b^2)$
$\therefore x=\frac{1}{2}log(a^2+b^2)$
To find 'y', divide (2) by (1) giving,
$b/a=tany$
$\therefore y=ta{n}^{-1}b/a$
$\therefore log(a+ib)=x+iy=\frac{1}{2}log(a^2+b^2)+ita{n}^{-1}\left(b/a\right)$