Question

log (a + ib ) when
a is greater than zero, b is less than zero

Answer 1

GIVEN :

log (a + ib ) when  a is greater than zero,and b is greater than zero

TO FIND :

The value of log(a+ib)

SOLUTION :

Given that Log (a+ib)

Let w=x+iy.

Log (a+ib)=w=x+iy.

∴ By laws of logarithm

$e^w=e^{x+iy}=e^{log(a+ib)}$

$e^{x+iy}=e^{log(a+ib)}$

By using the properties

$a^{m+n}=a^m.a^n$ and

$e^{logx}=x$

$e^x.e^{iy}=a+ib$

$e^x(cosy+isiny)=a+ib$

$e^xcosy+ie^xsiny=a+ib$

∴ equating the real & imaginary parts ,we get

$a=e^xcosy\hfill (1)$

$b=e^xsiny\hfill (2)$

Now to find  the values of x and y

Squaring the equations (1) & (2)

$a^2=e^{2x}cos^2y\hfill (3)$

$b^2=e^{2x}sin^2y\hfill (4)$

Now adding the equations (3) and (4)

$a^2+b^2=e^{2x}(cos^2y+sin^2y)$

By using the identity

$sin^2x+cos^2x=1$

$a^2+b^2=e^{2x}(1)$

$a^2+b^2=e^{2x}$

Rewritting the equation,

∴ $e^{2x}=a^2+b^2$  

Taking log on both sides  

$log(e^{2x})=log(a^2+b^2)$  

By using the property

$logx^n=nlogx$

$2xloge=log(a^2+b^2)$  

By using the property

$loge=1$

$2x(1)=log(a^2+b^2)$  

∴ $x=\frac{1}{2}log(a^2+b^2)$  

 Now divide the equation (2) by (1)

$\frac{b}{a}=\frac{e^xsiny}{e^xcosy}$

$\frac{b}{a}=\frac{siny}{cosy}$

$\frac{b}{a}=tany$

$tany=\frac{b}{a}$

∴ $y=tan^{-1}(\frac{b}{a})$

​Substitute the values of x and y

log(a+ib)=x+iy

$=\frac{1}{2}log(a^2+b^2)+itan^{-1}(\frac{b}{a})$

∴ the value is $log(a+ib)=\frac{1}{2}log(a^2+b^2)+itan^{-1}(\frac{b}{a})$