FINDING LOCUS OF A COMPLEX NUMBERS EXAMPLES

Finding Locus of a Complex Numbers Examples :

Here we are going to see some example problems based on finding locus of a complex numbers examples.

Finding Locus of a Complex Numbers Examples  - Questions

Question 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.

Solution :

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)=  1

√x2 + (y - 4)  =  √x2 + (y + 4)

Taking squares on both sides, we get

x2 + (y - 4)  =  x2 + (y + 4)

x2 + y2 - 2y(4) + 4  =  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.

Question 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

Solution :

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

Hence proved.

After having gone through the stuff given above, we hope that the students would have understood, "Finding Locus of a Complex Numbers Examples".

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