Question
Question: The locus represented by \( \left| {z - 1} \right| = \left| {z + i} \right| \) is: A. a circle of ...
The locus represented by ∣z−1∣=∣z+i∣ 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
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+iy , where i is the square root of -1 and the absolute value of a complex number, ∣x+iy∣ , is equal to x2+y2 . So in the given equation replace z with x+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 ∣z−1∣=∣z+i∣
Let z be x+iy
On replacing z in the equation ∣z−1∣=∣z+i∣ with x+iy , we get
∣x+iy−1∣=∣x+iy+i∣
Putting the terms without i together and with i together, we get
∣x−1+iy∣=∣x+i(y+1)∣
Write the absolute values of the above complex numbers
(x−1)2+y2=x2+(y+1)2
On squaring on both sides, we get
⇒(x−1)2+y2=x2+(y+1)2
We know the value of (a+b)2 is a2+2ab+b2 and the value of (a−b)2 is a2−2ab+b2
⇒x2−2x(1)+12+y2=x2+y2+2y(1)+12
⇒x2−2x+1+y2=x2+y2+2y+1
⇒−2x=2y
⇒2x+2y=0
⇒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=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.