Complex Numbers 3 Question 4
4. Let $z$ and $w$ be two complex numbers such that $|z| \leq 1$, $|w| \leq 1$ and $|z+i w|=|z-i \bar{w}|=2$, then $z$ equals
$(1995,2 M)$
(a) 1 or $i$
(b) $i$ or $-i$
(c) 1 or -1
(d) $i$ or -1
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Answer:
Correct Answer: 4. (c)
Solution:
- Given,
$$ |z+i w|=|z-i \bar{w}|=2 $$
$$ \begin{array}{ll} \Rightarrow & |z-(-i w)|=|z-(i \bar{w})|=2 \\ \Rightarrow & |z-(-i w)|=|z-(-i \bar{w})| \end{array} $$
$\therefore z$ lies on the perpendicular bisector of the line joining $-i w$ and $-i \bar{w}$. Since, $-i \bar{w}$ is the mirror image of $-i w$ in the $X$-axis, the locus of $z$ is the $X$-axis.
Let $\quad z=x+i y$ and $y=0$.
Now, $\quad|z| \leq 1 \Rightarrow x^{2}+0^{2} \leq 1 \Rightarrow-1 \leq x \leq 1$.
$\therefore z$ may take values given in option (c).
Alternate Solution
$$ \begin{aligned} & |z+i w| \leq|z|+|i w|=|z|+|w| \\ & \leq 1+1=2 \\ & \therefore \quad|z+i w| \leq 2 \\ & \Rightarrow \quad|z+i w|=2 \text { holds when } \\ & \arg z-\arg i w=0 \\ & \Rightarrow \quad \arg \frac{z}{i w}=0 \\ & \Rightarrow \frac{z}{i w} \text { is purely real. } \\ & \Rightarrow \frac{z}{w} \text { is purely imaginary. } \end{aligned} $$
Similarly, when $|z-i \bar{w}|=2$, then $\frac{z}{\bar{w}}$ is purely imaginary
Now, given relation
$$ |z+i w|=|z-i \bar{w}|=2 $$
Put $w=i$, we get
$$ \begin{aligned} \left|z+i^{2}\right| & =\left|z+i^{2}\right|=2 \\ |z-1| & =2 \end{aligned} $$
$$ \Rightarrow \quad z=-1 \quad[\because|z| \leq 1] $$
Put $w=-i$, we get
$$ \begin{aligned} & \Rightarrow \quad\left|z-i^{2}\right|=\left|z-i^{2}\right|=2 \\ & \Rightarrow \quad|z+1|=2 \Rightarrow z=1 \\ & \therefore z=1 \text { or }-1 \text { is the correct option. } \end{aligned} \quad[\because|z| \leq 1] $$
