Question

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

• A. The circle with radius $\frac{1}{2a}$ and centre $(\frac{1}{2a},0)$ for $a>0,b\ne 0$
• B. The circle with radius $-\frac{1}{2a}$ and centre $(-\frac{1}{2a},0)$ 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)

The Y - axis for $a=0,b\ne 0$

Here, $x+iy=\frac{1}{a+ibt}\times \frac{a-ibt}{a-ibt}\therefore x+iy=\frac{a-ibt}{a^2+b^2{t}^2}$
Let $a\ne 0,b\ne 0\therefore x=\frac{a}{a^2+b^2{t}^2}$
And $y=\frac{-bt}{a^2+b^2{t}^2}\Rightarrow \frac{y}{x}=\frac{-bt}{a}\Rightarrow t=\frac{-ay}{bx}$
On putting $x=\frac{a}{a^2+b^2{t}^2}$, we get
$x(a^2+b^2\cdot \frac{a^2{y}^2}{b^2{x}^2})=a\Rightarrow a^2(x^2+y^2)=ax$
Or $x^2+y^2-\frac{x}{a}=0$
Or ${(x-\frac{1}{2a})}^2+y^2=\frac{1}{4a^2}\therefore$ option (a) is correct.
For $a\ne 0$ and b=0,
$x+iy=\frac{1}{a}\Rightarrow x=\frac{1}{a},y=0\Rightarrow z$ lies on X - axis.
$\therefore$ option (c) is correct.
For $a=0$ and $b\ne 0,x+iy=\frac{1}{ibt}\Rightarrow x=0,y=-\frac{1}{bt}\Rightarrow z$ lies on Y - axis.
$\therefore$ option (d) is correct.