Question

If $\left|z+1\right|=\sqrt{2}\left|z-1\right|$, then the locus described by the point $z$ in the argand diagram is a

• A. straight line
• B. circle
• C. parabola
• D. none of these

Answer 1

The correct option is B

circle

Explanation for the correct option.

Find the locus of point $z$.

Let $z=x+iy$ be a complex number, then the equation $\left|z+1\right|=\sqrt{2}\left|z-1\right|$ can be written as:

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

Now, square both sides.

$$\begin{gathered} {\left(\sqrt{{\left(x+1\right)}^2+y^2}\right)}^2={\left(\sqrt{2}\sqrt{{\left(x-1\right)}^2+y^2}\right)}^2 \\ \Rightarrow {\left(x+1\right)}^2+y^2=2\left[{\left(x-1\right)}^2+y^2\right] \\ \Rightarrow x^2+2x+1+y^2=2\left(x^2-2x+1+y^2\right) \\ \Rightarrow 0=2x^2-4x+2+2y^2-x^2-2x-1-y^2 \\ \Rightarrow 0=x^2+y^2-6x+1 \end{gathered}$$

So point $z$ is defined by the line $x^2+y^2-6x+1=0$ and it is the equation of a circle.

So the locus described by the point $z$ in the argand diagram is a circle.

Hence, the correct option is B.