Question

The inequality $\left|z-4\right|$ less than $\left|z-2\right|$ represents the region given by

• A. $\text{Re(z)}$ greater than $0$
• B. $\text{Re(z)}$ less than $0$
• C. $\text{Re(z)}$ greater than $2$
• D. $\text{Re(z)}$ greater than $3$

Answer 1

The correct option is D

$\text{Re(z)}$ greater than $3$

Explanation for the correct option:

Solve the given inequality.

An inequality $\left|z-4\right|<\left|z-2\right|$ is given.

Assume that, $z=x+iy$.

So,

$$\begin{gathered} \left|x+iy-4\right|<\left|x+iy-2\right| \\ \Rightarrow \sqrt{{(x-4)}^2+y^2}<\sqrt{{(x-2)}^2+y^2} \\ \Rightarrow {(x-4)}^2+y^2<{(x-2)}^2+y^2 \\ \Rightarrow {(x-4)}^2<{(x-2)}^2 \\ \Rightarrow x^2+16-8x<x^2+4-4x \\ \Rightarrow 16-8x<4-4x \\ \Rightarrow 16-4<8x-4x \\ \Rightarrow 12<4x \\ \Rightarrow 3<x \end{gathered}$$

Therefore, the real part of $z$ is greater than $3$.

Hence, option $D$ is the correct answer.