Question

If $\left|z - 2 i\right|\leq \sqrt{2}$ , then the maximum value of $\left|3 + i \left(z - 1\right)\right|$ is

• A. $\sqrt{2}$
• B. $2\sqrt{2}$
• C. $2+\sqrt{2}$
• D. $3+2\sqrt{2}$

Answer 1

Answer: (B)

Given $\left|z - 2 i\right|\leq \sqrt{2}$ .
To find : maximum value of $\left|3 + i \left(z - 1\right)\right|$
$\therefore \left|3 + i \left(z - 1\right)\right|$
$=\left|3 + i z - i\right|$
$=\left|3 - i + i \left(z - 2 i\right) + 2 i^{2}\right|$
$=\left|1 - i + i \left(z - 2 i\right)\right|$ $\left[\because i^{2} = - 1\right]$
So, $\left|3 + i \left(z - 1\right)\right|\leq \left|1 - i\right|+\left|i \left(z - 2 i\right)\right|$ (using triangle inequality)
$\Rightarrow \left|3 + i \left(z - 1\right)\right|\leq \sqrt{1^{2} + \left(- 1\right)^{2}}+\sqrt{2}$
$\Rightarrow \left|3 + i \left(z - 1\right)\right|\leq 2\sqrt{2}$
Thus, the maximum value of $\left|3 + i \left(z - 1\right)\right|$ is $2\sqrt{2}$ .