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