Mathematics - Complex Number Question with Solution | TestHub

$x+iy=\frac{1}{a+ibt}$(Multiply & divide by conjugate$a+ibt$)
$x+iy=\frac{a-ibt}{a^2+b^2 t^2}$
Let$a \ne 0$and$b \ne 0$
$x=\frac{a}{a^2+b^2 t^2}$......(i)
$y=\frac{-bt}{a^2+b^2 t^2}$......(ii)
$\frac{y}{x}=\frac{-bt}{a} \Rightarrow t= -\frac{ay}{bx}$
Put in (i)
$x\left\{a^2+b^2.\frac{a^2{y}^2}{b^2{x}^2}\right\}=a$
$\Rightarrow$$a^2 \left(x^2+y^2\right)=ax$
$x^2+y^2-\frac{1}{a}x=0$
${\left(x-\frac{1}{2a}\right)}^2+y^2=\frac{1}{4a^2}$
Circle with centre$\left(\frac{1}{2a},0\right)$, radius$=\frac{1}{2\left|a\right|}$$\{\begin{array}{c}a>0\text{ ; }\frac{1}{2a}\\ a<0;\frac{-1}{2a}\end{array}$
For$a\ne 0, b=0$
$x+iy=\frac{1}{a}$
$x=\frac{1}{a}, y=0 \Rightarrow z$lies on x - axis
For$a=0, b\ne 0$
$x+iy=\frac{1}{ibt}$
$y= -\frac{1}{bt}, x=0$
$\Rightarrow$z lies on y - axis.