Complex Number Question 1
Question 1 - 2024 (01 Feb Shift 1)
LetS =${z \in C:|z-1|= \text{1 and} (\sqrt{2}-1)(z+\bar{z})-i(z-\bar{z})=2 \sqrt{2}}$. Let $z _1, \quad z _2 \in S$ 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 $(x-1)^{2}+y^{2}=1 \quad \rightarrow[$ 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 $x=1$ or $x=\frac{1}{2-\sqrt{2}} \rightarrow$
On solving (3) with (2)
we get
For $x=1 \Rightarrow y=1 \Rightarrow Z _2=1+i \&$ 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
$$ \begin{aligned} & \left|\sqrt{2} z _1-z _2\right|^{2} \\ & =\left|\left(\frac{1}{\sqrt{2}}+1\right) \sqrt{2}+i-(1+i)\right|^{2} \\ & =(\sqrt{2})^{2} \\ & =2 \end{aligned} $$
