Question

If $\omega =\frac{z}{z-\left({\displaystyle \frac{1}{3}}\right)i}$ and $\left|\omega \right|=1$, then $z$ lies on

• A. A parabola
• B. A straight line
• C. A circle
• D. An ellipse

Answer 1

The correct option is B

A straight line

Explanation for the correct option.

Step 1: Assumption and simplification required for the solution.

Let the complex number $z\text{be}x+iy$.

As given that $\left|\omega \right|=1$ and $\omega =\frac{z}{z-\left({\displaystyle \frac{1}{3}}\right)i}$

Now take mod on both sides of the equation $\omega =\frac{z}{z-\left({\displaystyle \frac{1}{3}}\right)i}$

$\left|\omega \right|=\left|\frac{z}{z-\left({\displaystyle \frac{1}{3}}\right)i}\right|$

This gives

$1=\left|\frac{z}{z-\left({\displaystyle \frac{1}{3}}\right)i}\right|$

Step 2: Calculation for the correct option

Since $z=x+iy$ then

$1=\frac{\left|x+iy\right|}{\left|x+iy-\left({\displaystyle \frac{1}{3}}\right)i\right|}$

Solving the above equation,

$\begin{array}{rcl}\therefore \left|x+iy-\left(\frac{1}{3}\right)i\right|& =& \left|x+iy\right|\\ & \Rightarrow & \left|3x+3iy-i\right|=\left|3x+3iy\right|\\ & \Rightarrow & \left|3x+\left(3y-1\right)i\right|={\left(3x\right)}^2+{\left(3iy\right)}^2\\ & \Rightarrow & {\left(3x\right)}^2+{\left(3y-1\right)}^2=9\left(x^2+y^2\right)\\ & \Rightarrow & 9x^2+1+9y^2-6y=9x^2+9y^2\\ & \Rightarrow & 6y-1=0\end{array}$

This is an equation of a straight line.

Hence, the correct option is (B).