Question

Let $a,b\in R$ and $a^2+b^2\ne 0$. Suppose $S=\{z\in C:z=\frac{1}{a+ibt},t\in R,t\ne 0\},$ 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 $(\frac{1}{2a},0)$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 correct options are
A 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 $a=0,b\ne 0$

$z=\frac{1}{a+ibt}\Rightarrow z=\frac{a-ibt}{a^2+b^2{t}^2}=x+iy\therefore x=\frac{a}{a^2+b^2{t}^2}\text{and}y=\frac{-bt}{a^2+b^2{t}^2}$

So assuming $a\ne 0$ and $b\ne 0$ $x^2+y^2=\frac{1}{a^2+b^2{t}^2}=\frac{x}{a}\Rightarrow x^2-\frac{x}{a}+y^2=0\Rightarrow {(x-\frac{1}{2a})}^2+y^2={\left(\frac{1}{2a}\right)}^2$

If $a\ne 0,b=0,\text{then}(x,y)=(\frac{1}{a},0)$

If $a=0,b\ne 0,\text{then}(x,y)=(0,\frac{-1}{bt})$