Question

If \[z = x + iy\] and \[\left| {z - zi} \right| = 1\],then [RPET\[1988,{\rm{ }}91\]]
A)z lies on x-axis
B)z lies on y-axis
C) z lies on circle
D) None of these

Answer 1

Hint: in this question we have to find where complex number x lies. Put z in form of real and imaginary number into the equation. First, write the z complex number as a combination of real and imaginary numbers. 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\[\left| {z - zi} \right| = 1\]
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\[\left| {z - zi} \right| = 1\]
\[\left| {(x + iy) - (x + iy)i} \right| = 1\]
We know that
\[\left| z \right| = \sqrt {{x^2} + {y^2}} \]
\[\sqrt {{{(x + y)}^2} + {{(y - x)}^2}} = 1\]
\[{(x + y)^2} + {(y - x)^2} = 1\]
\[{x^2} + {y^2} + 2xy + {y^2} + {x^2} - 2xy = 1\]
\[2{x^2} + 2{y^2} = 1\]
\[2({x^2} + {y^2}) = 1\]
This equation represents equation of circle.
Z lies on circle



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.