Question

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