Question

The number of complex numbers $z$ such that $\left|z-1\right|=\left|z+1\right|=\left|z-i\right|$ equal

• A. $0$
• B. $1$
• C. $2$
• D. Infinity

Answer 1

The correct option is B

$1$

Explanation for the correct option:

Step 1: Compute the required values of modulus of complex numbers.

Assume that, $z=x+iy$.

Compute the value of $\left|z-1\right|$

$$\begin{gathered} \left|z-1\right|=\left|x+iy-1\right| \\ \Rightarrow \left|z-1\right|=\left|\left(x-1\right)+iy\right| \\ \Rightarrow \left|z-1\right|=\sqrt{{(x-1)}^2+y^2} \end{gathered}$$

Compute the value of $\left|z+1\right|$

$$\begin{gathered} \left|z+1\right|=\left|x+iy+1\right| \\ \Rightarrow \left|z+1\right|=\left|\left(x+1\right)+iy\right| \\ \Rightarrow \left|z+1\right|=\sqrt{(x+1)+y^2} \end{gathered}$$

Compute the value of $\left|z-i\right|$

$$\begin{gathered} \left|z-i\right|=\left|x+iy+i\right| \\ \Rightarrow \left|z-i\right|=\left|x+i(y-1)\right| \\ \Rightarrow \left|z-i\right|=\sqrt{x^2+{(y-1)}^2} \end{gathered}$$

Step 2: Find the number of complex numbers satisfying the given condition.

It is given that, $\left|z-1\right|=\left|z+1\right|=\left|z-i\right|$.

Evaluate $\left|z-1\right|=\left|z+1\right|$ as follows:

$$\begin{gathered} \sqrt{{(x-1)}^2+y^2}=\sqrt{{(x+1)}^2+y^2} \\ \Rightarrow {(x-1)}^2+y^2={(x+1)}^2+y^2 \\ \Rightarrow x^2+1-2x+y^2=x^2+1+2x+y^2 \\ \Rightarrow -2x=2x \\ \Rightarrow -4x=0 \\ \Rightarrow x=0 \end{gathered}$$

Therefore, the value of $x$ is $0$.

Evaluate $\left|z-1\right|=\left|z-i\right|$ as follows:

$$\begin{gathered} \sqrt{{(x-1)}^2+y^2}=\sqrt{x^2+{(y-1)}^2} \\ \Rightarrow {(x-1)}^2+y^2=x^2+{(y-1)}^2\Rightarrow x^2+1-2x+y^2=x^2+y^2+1-2y \\ \Rightarrow -2x=-2y \end{gathered}$$

Since the value of $x$ is $0$.

So, $y=0$.

Therefore, the complex number $\mathrm{z}=0+0\mathrm{i}$ is the only complex number that satisfies the given condition.

Hence, option $B$ is the correct option.