Let z be a complex number such that the real part of z - 2 i z + 2 i is zero. Then, the maximum value of | z - ( 6 + 8 i ) | is equal to [2024]

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Solution

Q.
Let $z$ be a complex number such that the real part of $\frac{z - 2 i}{z + 2 i}$ is zero. Then, the maximum value of $| z - (6 + 8 i) |$ is equal to [2024]

1 $12$

2 $\infty$

3 $8$

4 $10$

Ans.
(1)

Let $z = x + i y, x, y \in R$

Consider $\frac{z - 2 i}{z + 2 i} = \frac{x + i y - 2 i}{x + i y + 2 i} = \frac{x + (y - 2) i}{x + (y + 2) i} \times \frac{x - (y + 2) i}{x - (y + 2) i}$

$= \frac{x^{2} + x (y + 2) i + x (y - 2) i + (y^{2} - 4)}{x^{2} + {(y + 2)}^{2}}$

$\Rightarrow \frac{x^{2} + y^{2} - 4}{x^{2} + {(y + 2)}^{2}} = 0 \Rightarrow x^{2} + y^{2} = 4$

$\Rightarrow | z |^{2} = 4 \Rightarrow | z | = 2$

Now, $| z - (6 + 8 i) | \leq | z | + | - 6 - 8 i | \leq 2 + 10 = 12$

$∴ | z - (6 + 8_{i}) |_{max} = 12$

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