Question

The locus of z given by\[\left| {\dfrac{{z - 1}}{{z - i}}} \right| = 1\], is [Roorkee\[1990\]]
A) A circle
B) An ellipse
C) A straight line
D) A Parabola

Answer 1

Hint: in this question we have to find what shape locus of the point in the complex plane represent. First, write the given complex number as a combination of real and imaginary numbers. Put z in the form of real and imaginary numbers 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: A complex number equation\[\left| {\dfrac{{z - 1}}{{z - i}}} \right| = 1\]
Now we have complex number \[\left| {\dfrac{{z - 1}}{{z - i}}} \right| = 1\]
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\[\left| {\dfrac{{z - 1}}{{z - i}}} \right| = 1\]
\[\left| {\dfrac{{(x + iy) - 1}}{{(x + iy) - i}}} \right| = 1\]
\[\left| {(x + iy) - 1} \right| = \left| {(x + iy) - i} \right|\]
We know that
\[\left| z \right| = \sqrt {{x^2} + {y^2}} \]
\[\sqrt {{{(x - 1)}^2} + {y^2}} = \sqrt {{x^2} + {{(y - 1)}^2}} \]
\[{x^2} + 1 - 2x + {y^2} = {x^2} + {y^2} + 1 - 2y\]
\[2x = 2y\]
\[x = y\]
\[x - y = 0\]
This equation represents equation of line.
Here \[x - y = 0\]represent the equation of line therefore locus of point represent line.



Option ‘C’ 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.