FINDING LOCUS OF COMPLEX NUMBERS

Finding Locus of Complex Numbers :

To find the locus of a complex number, first we have to replace z by the complex number x + iy and simplify.

Let us look into some problems to understand the concept

Finding Locus of Complex Numbers - Problems

Obtain the Cartesian form of the locus of z = x + iy in each of the following cases:

(i)  [Re (iz)]²  =  3

Solution :

z = x + iy

iz  =  i(x + iy)

    =   ix + i²y

iz   =  ix - y

Real part  =  y

[Re (iz)]²  =  3

  y²  =  3

  y² - 3  =  0

(ii)  im [(1 - i)z + 1]  =  0

Solution :

z = x + iy

(1 - i)z + 1  =  (1 - i)(x + iy) + 1

  =  x + iy - ix - i²y + 1

  =  x + iy - ix + y + 1

  =  (x + y + 1) + i(y - x)

imaginary part  =  y - x

y - x  =  0

x - y  =  0

Hence x - y  =  0 is the locus.

(iii)  |z + i|  =  |z - 1|

Solution :

z = x + iy

|z + i|  =  |x + iy + i|

|z + i|  =  |x + i(y + 1)|  ----(1)

|z - 1|  =  |x + iy - 1|

|z - 1|  =  |(x - 1) + iy| ----(2)

(1)  =  (2)

|x + i(y + 1)|  =  |(x - 1) + iy|

√x² + (y + 1)²  =  √(x - 1)² + y²

Taking squares on both sides, we get

x² + (y + 1)²  =  (x - 1)² + y²

x² + y² + 2y + 1  =  x² - 2x + 1 + y²

2x + 2y + 1 - 1   =  0

x + y  =  0

Hence the locus is x + y  =  0.

(iv)  z bar  =  z^-1

Solution :

z = x + iy

L.H.S :

z bar  =  (x + iy) bar

z bar  =  x - iy  -----(1)

R.H.S :

z^-1  =  1/z  =  1/(x + iy)

By multiplying the conjugate of the denominator, we get

1/(x + iy)  =  (x - iy)/(x + iy)(x - iy)

  =  (x - iy) / (x² + y²) --------(2)

L.H.S  =  R.H.S

(1)  =  (2)

x - iy  =  (x - iy) / (x² + y²)

1  =  1/ (x² + y²)

x² + y²  =  1

After having gone through the stuff given above, we hope that the students would have understood how to find locus of a complex number.

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