Complex Numbers Ques 88
- Let $a, b \in R$ and $a^{2}+b^{2} \neq 0$.
Suppose $S=z \in C: z=\frac{1}{a+i b t}, t \in R, t \neq 0$, where $i=\sqrt{-1}$. If $z=x+i y$ and $z \in S$, then $(x, y)$ lies on
(2016 Adv.)
(a) the circle with radius $\frac{1}{2 a}$ and centre $\frac{1}{2 a}, 0$ for $a>0, b \neq 0$
(b) the circle with radius $-\frac{1}{2 a}$ and centre $-\frac{1}{2 a}, 0$ for $a<$ $0, b \neq 0$
(c) the $X$-axis for $a \neq 0, b=0$
(d) the $Y$-axis for $a=0, b \neq 0$
Show Answer
Answer:
Correct Answer: 88.(d)
Solution:
- Here, $x+i y=\frac{1}{a+i b t} \times \frac{a-i b t}{a-i b t}$
$\therefore \quad x+i y=\frac{a-i b t}{a^{2}+b^{2} t^{2}}$
Let $\quad a \neq 0, b \neq 0$
$\therefore \quad x=\frac{a}{a^{2}+b^{2} t^{2}}$ and $y=\frac{-b t}{a^{2}+b^{2} t^{2}}$
$\Rightarrow \quad \frac{y}{x}=\frac{-b t}{a} \Rightarrow t=\frac{a y}{b x}$
On putting $x=\frac{a}{a^{2}+b^{2} t^{2}}$, we get
$$ x a^{2}+b^{2} \cdot \frac{a^{2} y^{2}}{b^{2} x^{2}}=a \Rightarrow a^{2}\left(x^{2}+y^{2}\right)=a x $$
or $\quad x^{2}+y^{2}-\frac{x}{a}=0$
or
$$ x-\frac{1}{2a}^{2}+y^{2}=\frac{1}{4a^{2}} $$
$\therefore$ Option (a) is correct.
For
$$ \begin{gathered} a \neq 0 \text { and } b=0, \\ x + iy = \frac{1}{a} \Rightarrow x = \frac{1}{a}, y = 0 \end{gathered} $$
$\Rightarrow z$ lies on $X$-axis.
$\therefore$ Option (c) is correct.
For $a=0$ and $b \neq 0, x+i y=\frac{1}{i b t} \Rightarrow x=0, y=-\frac{1}{b t}$
$\Rightarrow z$ lies on $Y$-axis.
$\therefore$ Option $(d)$ is correct.
BETA
