Question

If \[z = x + iy\] and \[|z - 2 + i| = |z - 3 - i|\]then locus of z is [RPET \[1999\]]
A) \[2x + 4y - 5 = 0\]
B) \[2x - 4y - 5 = 0\]
C) \[x + 2y = 0\]
D) \[x - 2y + 5 = 0\]

Answer 1

Hint: in this question we have to find the locus of complex number z. First, write the given complex number as a combination of real and imaginary numbers. Then equate the imaginary part of the complex number with the given value.



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 \[|z - 2 + i| = |z - 3 - i|\]
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 \[|z - 2 + i| = |z - 3 - i|\]
\[|(x + iy) - 2 + i| = |(x + iy) - 3 - i|\]
\[|(x - 2) + i(y + 1)| = |(x - 3 + i(y - 1)|\]
We know that
\[\left| z \right| = \sqrt {{x^2} + {y^2}} \]
Now equation \[|(x - 2) + i(y + 1)| = |(x - 3 + i(y - 1)|\]is written as
\[\sqrt {{{(x - 2)}^2} + {{(y + 1)}^2}} = \sqrt {x - 3{)^2} + {{(y - 1)}^2}} \]
\[{(x - 2)^2} + {(y + 1)^2} = x - 3{)^2} + {(y - 1)^2}\]
\[{x^2} + 4 - 4x + {y^2} + 1 + 2y = {x^2} + 9 - 6x + {y^2} + 1 - 2y\]
Now locus of complex number z is \[2x + 4y - 5 = 0\]



Option ‘A’ 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.
Don’t try to put the value of z in a given equation first try to simplify it after that value of z would be put in the equation.