Question

If \[w = \dfrac{z}{{z - \dfrac{1}{3}i}}\] and \[\left| w \right| = 1\], then z lies on [AIEEE\[2005\]]
A) A straight line
B) A parabola
C) An ellipse
D) A circle

Answer 1

Hint: in this question we have to find where given complex number lies which satisfy given condition. First write the given complex number as a combination of real and imaginary number. Put z in form of real and imaginary number into the equation.

Formula Used: Equation of complex number is given by
\[z = x + iy\]
Where
z is a complex number
x represent real part of complex number
iy is a imaginary part of complex number
i is iota
Square of iota is equal to the negative of one

Complete step by step solution: Given: Equation in the form of complex number
Now we have complex number equation\[w = \dfrac{z}{{z - \dfrac{1}{3}i}}\]
\[\left| {\dfrac{z}{{z - \dfrac{1}{3}i}}} \right| = 1\]
\[\left| z \right| = \left| {z - \dfrac{1}{3}i} \right|\]
We know that complex number is written as a combination of real and imaginary number.
\[z = x + iy\]
Where
z is a complex number
x represent real part of complex number
iy is a imaginary part of complex number
Put this value in \[\left| z \right| = \left| {z - \dfrac{1}{3}i} \right|\]
\[\left| {x + iy} \right| = \left| {(x + iy) - \dfrac{1}{3}i} \right|\]
\[\left| {x + iy} \right| = \left| {x + i(y - \dfrac{1}{3})} \right|\]
We know that
\[\left| z \right| = \sqrt {{x^2} + {y^2}} \]
\[\sqrt {{x^2} + {y^2}} = \sqrt {{x^2} + {{(y - \dfrac{1}{3})}^2}} \]
\[{x^2} + {y^2} = {x^2} + {(y - \dfrac{1}{3})^2}\]
\[{y^2} = {(y - \dfrac{1}{3})^2}\]
\[{y^2} = {y^2} + \dfrac{1}{9} - \dfrac{2}{3}y\]
\[\dfrac{2}{3}y = \dfrac{1}{9}\]
\[y = \dfrac{1}{6}\]
This equation represents equation of line.
So z lies on straight line.

Option ‘A’ is correct

Note: Complex number is a number which is a combination of real and imaginary number. So in complex number questions, we have to represent number as a combination of real and its imaginary part. Imaginary part is known as iota. Square of iota is equal to negative one.