**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.

**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

√x^{2} + (y - 4)^{2 }/ √x^{2} + (y + 4)^{2 }= 1

√x^{2} + (y - 4)^{2 } = √x^{2} + (y + 4)^{2 }

Taking squares on both sides, we get

x^{2} + (y - 4)^{2 } = x^{2} + (y + 4)^{2 }

x^{2} + y^{2} - 2y(4) + 4^{2 } = x^{2} + y^{2} + 2y(4) + 4^{2}

x^{2} + y^{2} - 8y + 16^{ } = x^{2} + y^{2} + 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 2x^{2} + 2y^{2} + 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 + i ^{2}y + 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} - x^{2}] = 0**

**-2x ^{2} - x + 2y - 2y^{2} = 0**

Multiply through out the equation by negative, we get

**2x ^{2} + x - 2y + 2y^{2} = 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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