Question

Let $a,b\in R$ and $a^2+b^2\ne 0$. Suppose $S=\left\{z\in C:z=\frac{1}{a+ibt},t\in R,t\ne 0\right\}$, where $i=\sqrt{-1}$ If $z=x+iy$ and $z\in S$, then $(x,y)$ lies on

• A. the circle with radius $\frac{1}{2a}$ and centre $\left(\frac{1}{2a},0\right)$ for $a>0,b\ne 0$
• B. the circle with radius $-\frac{1}{2a}$ and centre $\left(-\frac{1}{2a},0\right)$ for $a>0,b\ne 0$
• C. the $x$-axis for $a\ne 0,b=0$
• D. the $y$-axis for $a=0,b\ne 0$

Answer 1

Answer: (A), (C), (D)

$z=x+iy=\frac{1}{a+ibt}=\frac{a-ibt}{a^2+b^2{t}^2}$
$x=\frac{a}{a^2+b^2{t}^2},y=\frac{-bt}{a^2+b^2{t}^2}$
$\therefore \frac{y}{x}=\frac{-bt}{a}$
$\Rightarrow t=\frac{-ay}{bx}$
$Sox\left(a^2+b^2\frac{a^2{y}^2}{b^2{x}^2}\right)=a$
$a^2{x}^2+a^2{y}^2=ax$
$x^2+y^2=\frac{x}{a}$
$\Rightarrow {\left(x-\frac{1}{2a}\right)}^2+y^2=\frac{1}{4a^2}$
circle with centre $\left(\frac{1}{2a},0\right)$ and radius $=\frac{1}{2a}$
for $a>0,b\ne 0$
If $b=0,a\ne 0$
$x+iy=\frac{1}{a}$
$\Rightarrow x=\frac{1}{a},y=0$
So $x$-axis
If $a=0,b\ne 0$
$x+iy=\frac{1}{ibt}$
$\Rightarrow x=0;y=-\frac{1}{bt}$
So y-axis