Question

Locus of the point z satisfying the equation \[\left| {iz - 1} \right| + \;\left| {z - i} \right| = 2\;\]is [Roorkee\[1999\]]
A) A straight line
B) A circle
C) An ellipse
D) A pair of straight lines

Answer 1

Hint: in this question we have to find what shape locus of the point in the complex plane represents. First, write the z complex number as a combination of real and imaginary numbers. Put z in the form of real and imaginary numbers 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\[\left| {iz - 1} \right| + \;\left| {z - i} \right| = 2\;\]
We know that complex numbers are written as a combination of real and imaginary numbers.
\[z = x + iy\]
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\[\left| {iz - 1} \right| + \;\left| {z - i} \right| = 2\;\]
\[\left| {i(x + iy) - 1} \right| + \;\left| {(x + iy) - i} \right| = 2\;\]
\[\left| { - (y + 1) + ix} \right| + \;\left| {x + i(y - 1)} \right| = 2\;\]
We know that
\[\left| z \right| = \sqrt {{x^2} + {y^2}} \]
\[\sqrt {{{( - (y + 1))}^2} + {x^2}} + \sqrt {{x^2} + {{(y - 1)}^2}} = 2\]
\[\sqrt {{{(y + 1)}^2} + {x^2}} = 2 - \sqrt {{x^2} + {{(y - 1)}^2}} \]
\[{y^2} + 1 + 2y + {x^2} = 4 + {x^2} + {y^2} + 1 - 2y - 4\sqrt {{x^2} + {{(y - 1)}^2}} \]
\[4y = 4 - 4\sqrt {{x^2} + {{(y - 1)}^2}} \]
\[y = 1 - \sqrt {{x^2} + {{(y - 1)}^2}} \]
\[{x^2} + {(y - 1)^2} = {(1 - y)^2}\]
\[{x^2} + {y^2} + 1 - 2y = 1 + {y^2} - 2y\]
\[{x^2} = 0\]
\[x = 0\]
This equation represents an equation of line.
Here \[x = 0\]represents the equation of line therefore the locus of point represents the line.



Option ‘A’ is correct



Note: Complex number is a number which is a combination of real and imaginary numbers. So in complex number questions, we have to represent the number as a combination of real and its imaginary part. Imaginary part is known as iota. Square of iota is equal to the negative one.