Finding the Locus of a Complex Number :
In this section, we will learn how to find the locus of a complex numbers through the following examples.
Example 1 :
If z = x + iy is a complex number such that |(z - 4i)/(z + 4i)| = 1 show that the locus of z is real axis.
Given that :
|(z - 4i)/(z + 4i)| = 1
|z - 4i|/|z + 4i| = 1
z = x + iy
|x + iy - 4i| / |x + iy + 4i| = 1
|x + i(y - 4)| / |x + i(y + 4)| = 1
√x2 + (y - 4)2 / √x2 + (y + 4)2 = 1
√x2 + (y - 4)2 = √x2 + (y + 4)2
Taking squares on both sides, we get
x2 + (y - 4)2 = x2 + (y + 4)2
x2 + y2 - 2y(4) + 42 = x2 + y2 + 2y(4) + 42
x2 + y2 - 8y + 16 = x2 + y2 + 8y + 16
-8y - 8y = 0
-16y = 0
y = 0
Real axis means, imaginary part must be zero in the complex plane. In x + iy, x is the real part and y is the imaginary part.
Since the value of y is 0, we have shown that locus z is real axis.
Example 2 :
If z = x + iy is a complex number such that im (2z + 1)/(iz + 1) = 0, show that locus of z is 2x2 + 2y2 + x - 2y = 0
im (2z + 1)/(iz + 1) = 0
z = x + iy
2z + 1 = 2(x + iy) + 1
= 2x + i2y + 1
= (2x + 1) + i2y ---(1)
iz + 1 = i(x + iy) + 1
= ix + i2y + 1
= ix - y + 1
= (1 - y) + ix ---(2)
(1) / (2)
(2z + 1) / (iz + 1) = [(2x + 1) + i2y] / [(1 - y) + ix]
By multiplying the conjugate of the denominator, we get
= [(2x+1) + i2y] [(1-y) - ix] / [(1 - y) + ix] [(1 - y) - ix]
im [(2z + 1) / (iz + 1)] = 0
[-x(2x + 1) + 2y(1 - y)] / [(1 - y)2 - x2] = 0
-2x2 - x + 2y - 2y2 = 0
Multiply through out the equation by negative, we get
2x2 + x - 2y + 2y2 = 0
After having gone through the stuff given above, we hope that the students would have understood how to find the locus of a complex number.
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