Question

The region of the complex plane for which \[\left| {\dfrac{{z - a}}{{z + \overline a }}} \right| = 1\] [\[R\left( a \right) \ne 0\]] is
A) x-axis
B) y-axis
C) A straight line \[x = a\]
D) None of these

Answer 1

Hint: in this question we have to find the region of complex plane in which given condition satisfy. Just simplify the given equation and use properties of conjugate of complex number to get equation.



Formula Used:\[z = x + iy\]
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: Complex equation with condition
Now we have equation\[\left| {\dfrac{{z - a}}{{z + \overline a }}} \right| = 1\]
\[\left| {z - a} \right| = \left| {z + \overline a } \right|\]
\[{\left| {z - a} \right|^2} = {\left| {z + \overline a } \right|^2}\]
\[(z - a)(\overline {z - a} ) = (z + \overline a )(\overline {z + \overline a } )\]
\[(z - a)(\overline z - \overline a ) = (z + \overline a )(\overline z + a)\]
\[z\overline z - z\overline a - a\overline z + a\overline a = z\overline z + za + \overline a \overline z + \overline a a\]
\[za + z\overline a + \overline a \overline z + a\overline z = 0\]
\[(a + \overline a )(z + \overline z ) = 0\]
\[(z + \overline z ) = 0\] (Because \[a + \overline a = 2Re(a) \ne 0\])
\[2Re(z) = 0\]
\[2x = 0\]
\[x = 0\] it is a equation of y-axis



Option ‘B’ is correct



Note: Complex number is a number which is a combination of real and imaginary number. So in combination number question 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.