Question

If \[z = x + iy\]and\[w = \left( {\dfrac{{1 - iz}}{{z - i}}} \right)\], then \[\left| w \right| = 1\] implies that, in the complex plane
A. z lies on the imaginary axis
B. z lies on the real axis
C. z lies on the unit circle
D. None of these

Answer 1

Hint: We substitute the value of ‘w’ in the equation \[\left| w \right| = 1\] and break the modulus function to form an equation having modulus on both sides of the equation. Use the value of ‘z’ to find the value of ‘iz’ and substitute those values in the equation.
* In a complex number \[z = x + iy\], the real part is x and the imaginary part is y.
* \[{i^2} = - 1\]
* A complex number is said to be purely real if it has an imaginary part equal to zero and purely imaginary if it has a real part equal to zero.
* Conjugate of a complex number \[z = x + iy\] is denoted by \[\overline z \] and is given by taking the negative value of the imaginary part i.e. \[\overline z = x - iy\].
* Modulus value of \[z = x + iy\] is given by \[\left| z \right| = \sqrt {{x^2} + {y^2}} \]

Complete step-by-step solution:
We are given that\[w = \left( {\dfrac{{1 - iz}}{{z - i}}} \right)\].
Since \[\left| w \right| = 1\],,,,,,,,,,,,,,,,,,,… (1)
Substitute the value of \[w = \left( {\dfrac{{1 - iz}}{{z - i}}} \right)\] in equation (1)
\[ \Rightarrow \left| {\dfrac{{1 - iz}}{{z - i}}} \right| = 1\]
We can break modulus in fraction to numerator and denominator
\[ \Rightarrow \dfrac{{\left| {1 - iz} \right|}}{{\left| {z - i} \right|}} = 1\]
Cross multiply the values from denominator in LHS to RHS
\[ \Rightarrow \left| {1 - iz} \right| = \left| {z - i} \right|\]........................… (2)
We are given \[z = x + iy\]...................… (3)
To find the value of ‘iz’ we multiply ‘z’ by ‘i’
\[ \Rightarrow iz = i(x + iy)\]
Open the terms in RHS
\[ \Rightarrow iz = ix + {i^2}y\]
Substitute the value of \[{i^2} = - 1\]in RHS
\[ \Rightarrow iz = ix - y\]
\[ \Rightarrow iz = - y + ix\] …………………..… (4)
Substitute the values from equations (3) and (4) in equation (2)
\[ \Rightarrow \left| {1 - ( - y + ix)} \right| = \left| {(x + iy) - i} \right|\]
\[ \Rightarrow \left| {1 + y + ix} \right| = \left| {x + iy - i} \right|\]
\[ \Rightarrow \left| {(1 + y) + ix} \right| = \left| {x + i(y - 1)} \right|\].................… (5)
Use the formula of modulus on both sides of the equation
\[ \Rightarrow \sqrt {{{(1 + y)}^2} + {{(x)}^2}} = \sqrt {{{(x)}^2} + {{(y - 1)}^2}} \]
Square both sides of the equation
\[ \Rightarrow {\left( {\sqrt {{{(1 + y)}^2} + {{(x)}^2}} } \right)^2} = {\left( {\sqrt {{{(x)}^2} + {{(y - 1)}^2}} } \right)^2}\]
Cancel square root by square power on both sides of the equation
\[ \Rightarrow {(1 + y)^2} + {(x)^2} = {(x)^2} + {(y - 1)^2}\]
Cancel same terms from both sides of the equation
\[ \Rightarrow {(1 + y)^2} = {(y - 1)^2}\]
Use identity \[{(a + b)^2} = {a^2} + {b^2} + 2ab\]in LHS and \[{(a - b)^2} = {a^2} + {b^2} - 2ab\] in RHS
\[ \Rightarrow {1^2} + {y^2} + 2 \times 1 \times y = {1^2} + {y^2} - 2 \times 1 \times y\]
\[ \Rightarrow {1^2} + {y^2} + 2y = {1^2} + {y^2} - 2y\]
Cancel same terms from both sides of the equation
\[ \Rightarrow 2y = - 2y\]
Shift all values to LHS of the equation
\[ \Rightarrow 2y + 2y = 0\]
\[ \Rightarrow 4y = 0\]
Divide both sides by 4
\[ \Rightarrow y = 0\]
So, we have y coordinate as zero.
Since we know a complex number is said to be purely real if it has an imaginary part equal to zero.
Here imaginary part (y) of complex number ‘z’ is zero
\[ \Rightarrow \]Complex numbers are purely real.
\[ \Rightarrow \]Z lies on the real line.

\[\therefore \]Option B is correct.

Note: Students are likely to make mistake of not substituting the value of \[{i^2} = - 1\] in the solution and they try to group together the terms for imaginary part and they include the value with \[{i^2}\] in it, which is wrong. Keep in mind that value of \[{i^2} = - 1\] so it is completely real.
Also, some students write the value of the imaginary part with i along it which should be taken care of, as the imaginary part is the value except i.