Complex Number Question 1
Question 1 - 2024 (01 Feb Shift 1)
LetS $={z \in C:|z-1|=1$ and $(\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2}}$. Let $\mathrm{z}{1}, \quad \mathrm{z}{2} \in S$ be such that
$\left|z_{1}\right|=\max {z \in s}|z|$ and $\left|z{2}\right|=\min {z \in S}|z|$. Then $\left|\sqrt{2} z{1}-z_{2}\right|^{2}$ equals :
(1) 1
(2) 4
(3) 3
(4) 2
Show Answer
Answer (4)
Solution
Let $Z=x+$ iy
Then $(\mathrm{x}-1)^{2}+\mathrm{y}^{2}=1 \quad \rightarrow[$ OPTION 1$]$
$& \quad(\sqrt{2}-1)(2 x)-i(2 i y)=2 \sqrt{2}$
$\Rightarrow(\sqrt{2}-1) x+y=\sqrt{2} \rightarrow(2)$
Solving (1) & (2)
we get
Either $\mathrm{x}=1$ or $x=\frac{1}{2-\sqrt{2}} \rightarrow$
On solving (3) with (2)
we get
For $\mathrm{x}=1 \Rightarrow \mathrm{y}=1 \Rightarrow \mathrm{Z}_{2}=1+\mathrm{i} &$ for
$x=\frac{1}{2-\sqrt{2}} \Rightarrow y=\sqrt{2}-\frac{1}{\sqrt{2}} \Rightarrow Z_{1}=\left(1+\frac{1}{\sqrt{2}}\right)+\frac{i}{\sqrt{2}}$
Now
$$ \begin{aligned} & \left|\sqrt{2} z_{1}-z_{2}\right|^{2} \ & =\left|\left(\frac{1}{\sqrt{2}}+1\right) \sqrt{2}+i-(1+i)\right|^{2} \ & =(\sqrt{2})^{2} \ & =2 \end{aligned} $$
