Solveeit Logo

Question

Question: The locus represented by \( \left| {z - 1} \right| = \left| {z + i} \right| \) is: A. a circle of ...

The locus represented by z1=z+i\left| {z - 1} \right| = \left| {z + i} \right| is:
A. a circle of radius 1 unit
B. An ellipse with foci at (1, 0) and (0, 1)
C. A straight line through the origin
D. A circle on the line joining (1, 0) and (0, 1) as diameter

Explanation

Solution

Hint : z is a complex number. A complex number has both a real part and an imaginary part. A complex number is represented in an ordered pair in the complex plane. The general form every complex number, z, is z=x+iyz = x + iy , where i is the square root of -1 and the absolute value of a complex number, x+iy\left| {x + iy} \right| , is equal to x2+y2\sqrt {{x^2} + {y^2}} . So in the given equation replace z with x+iyx + iy , and find its absolute values to find the locus.

Complete step by step solution:
We are given to find the locus represented by the line z1=z+i\left| {z - 1} \right| = \left| {z + i} \right|
Let z be x+iyx + iy
On replacing z in the equation z1=z+i\left| {z - 1} \right| = \left| {z + i} \right| with x+iyx + iy , we get
x+iy1=x+iy+i\left| {x + iy - 1} \right| = \left| {x + iy + i} \right|
Putting the terms without i together and with i together, we get
x1+iy=x+i(y+1)\left| {x - 1 + iy} \right| = \left| {x + i\left( {y + 1} \right)} \right|
Write the absolute values of the above complex numbers
(x1)2+y2=x2+(y+1)2\sqrt {{{\left( {x - 1} \right)}^2} + {y^2}} = \sqrt {{x^2} + {{\left( {y + 1} \right)}^2}}
On squaring on both sides, we get
(x1)2+y2=x2+(y+1)2\Rightarrow {\left( {x - 1} \right)^2} + {y^2} = {x^2} + {\left( {y + 1} \right)^2}
We know the value of (a+b)2{\left( {a + b} \right)^2} is a2+2ab+b2{a^2} + 2ab + {b^2} and the value of (ab)2{\left( {a - b} \right)^2} is a22ab+b2{a^2} - 2ab + {b^2}
x22x(1)+12+y2=x2+y2+2y(1)+12\Rightarrow {x^2} - 2x\left( 1 \right) + {1^2} + {y^2} = {x^2} + {y^2} + 2y\left( 1 \right) + {1^2}
x22x+1+y2=x2+y2+2y+1\Rightarrow {x^2} - 2x + 1 + {y^2} = {x^2} + {y^2} + 2y + 1
2x=2y\Rightarrow - 2x = 2y
2x+2y=0\Rightarrow 2x + 2y = 0
x+y=0\Rightarrow x + y = 0
Therefore, the above resulted line is a straight line. When we substitute x as 0 then y will also be 0. This means that the line x+y=0x + y = 0 is a straight line and passes through origin.
So, the correct answer is “Option C”.

Note : The absolute value of z is the distance of z from 0 in the complex plane. As the complex number can be represented in an ordered pair (x, y), distance from point (x, y) to 0 (0, 0) is the absolute value of z. Real numbers are all the numbers on the number line but complex numbers have a plane with an additional real axis to calculate the square roots of negative numbers.