Test Index

Complex Numbers Part 2

Section: Mathematics

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Question : 51 of 106

Marks: +1, -0

Let S

=\{z \in C: |z-1| = 1

(\sqrt{2} -1)(z+ \overline{z}) -i(z-\overline{z}) = 2 {\sqrt{2} \}}

z_1, z_2 \in S

|z_1| = \underset{z \in S}{\max} |z|

z_2 = \underset{z \in S}{\min} |z|

|\sqrt{2} z_1 - z_2|^2

[1-Feb-2024 Shift 1]

1
4
3
2

Solution: 👈: Video Solution

Let

Z = x + i y

Then

(x-1)^2 + y^2 = 1

→ (1)

(\sqrt{2}-1)(2x) - i(2i 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

(3)

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

|\sqrt{2} z_1 - z_2|^2

= \left| \left(\frac{1}{\sqrt{2}} + 1\right) \sqrt{2} + i - (1+i) \right|^2

= (\sqrt{2})^2

= 2

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