Question

If \[\omega = \dfrac{{1 - iz}}{{z - i}}\] and \[|\omega | = 1\]then show that in complex plane \[RPET{\rm{ }}1985,{\rm{ }}97;{\rm{ }}IIT{\rm{ }}1983;{\rm{ }}DCE{\rm{ }}2000,{\rm{ }}01;{\rm{ }}UPSEAT{\rm{ }}2003;{\rm{ }}MP{\rm{ }}PET{\rm{ }}2004\]]
A) z will be at imaginary axis
B) z will be at real axis
C) z will be at unity circle
D) None of these

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 and \[|\omega | = 1\]
Now we have complex number equation\[\omega = \dfrac{{1 - iz}}{{z - i}}\]
\[|\omega | = 1\]
\[|\dfrac{{1 - iz}}{{z - i}}| = 1\]
\[|1 - iz| = \left| {z - i} \right|\]
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
i is iota
Put this value in\[|1 - iz| = \left| {z - i} \right|\]
\[|1 - i(x + iy)| = \left| {(x + iy) - i} \right|\]
\[|(1 + y) - ix| = \left| {x + i(y - 1)} \right|\]
We know that
\[\left| z \right| = \sqrt {{x^2} + {y^2}} \]
\[\sqrt {{x^2} + 1 + {y^2} + 2y} = \sqrt {{x^2} + {y^2} + 1 - 2y} \]
\[{x^2} + 1 + {y^2} + 2y = {x^2} + {y^2} + 1 - 2y\]
\[4y = 0\]
\[y = 0\]
This is equation of straight line.
So z lies on real axis.

Option ‘B’ 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.