## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.6

Question 1.

If 2 = x + iy is a complex number such that \(\left|\frac{z-4 i}{z+4 i}\right|\) = 1 show that the locus of z is real axis.

Solution:

Let z = x + iy

\(\left|\frac{z-4 i}{z+4 i}\right|\) = 1 ⇒ |z – 4i| = |z + 4i|

⇒ |x + iy – 4i| = |x + iy + 4i|

⇒ |x + (y – 4)i| = |x + (y + 4)i|

x² + (y – 4)² = x² + (y + 4)²

⇒ (y – 4)² = (y + 4)² = ⇒ y – 4 = ±(y + 4)

y – 4 = -y – 4

2y – 0; y – 0 ⇒ x-axis

The locus is real axis

Samacheer Kalvi 12th Maths Guide

Question 2.

If z = x + iy is a complex number such that Im \(\left(\frac{2 z+1}{i z+1}\right)\) = 0, show that the locus of z is 2x² + 2y² + x – 2y = 0.

Solution:

Let z = x + iy

⇒ 2y – 2y² – 2x² – x = 0

⇒ 2x² + 2y² – 2y + x = 0

Question 3.

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

(i) [Re (iz)]² = 3

Solution:

since x + iy = z

iz = ix – y

∴ Re (iz) = -y

⇒ (y)² = 3

⇒ y² – 3 = 0

y² = 3

(ii) Im[(1 – i)z + 1] = 0

Solution:

Im ((1 – i) (x + iy) + 1) = 0

Im [x – ix + iy + y + 1] = 0

⇒ -x + y = 0 or x – y = 0

(iii) |z + i| = |z – 1|

Solution:

|x + iy + i| = |x + (y – 1)|

|x + i(y + 1)| = |x – 1) + iy

⇒ x² + (y + 1)² = (x – 1)² + y²

x² + y² + 2y + 1 = x² – 2x + 1 + y²

⇒ 2x + 2y = 0

x + y = 0

(iv) \(\bar {z}\) = z^{-1}

Solution:

x² + y² = 1, x² + y² = -1 which cannot be true.

∴ x² + y² = 1

Question 4.

Show that the following equations represent a circle, and, find its centre and radius.

(i) |z – 2 – i| = 3

Solution:

Let z = x + iy

|z – 2 – i| = 3 ⇒ |z – (2 + i)| = 3

∴ It is a circle with centre at 2 + i

i.e., (2, 1) and radius = 3 units

(ii) |2z + 2 – 4i| = 2

Solution:

2 |z + 1 – 2i| = 2

⇒|z – (-1 + 2i)| = 1

⇒ It is a circle with centre at (-1 + 2i)

⇒ (-1, 2) and radius = 1 unit

(iii) |3z – 6 + 12i| = 8.

Solution:

3 |z – 2 + 4i| = 8

⇒ |z – (2 – 4i)| = \(\frac{8}{3}\)

It is a circle with centre at (2 – 4i) ⇒ (2, -4) and radius \(\frac{8}{3}\) units.

Question 5.

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

(i) |z – 4| = 16

Solution:

Let z = x + iy

|x + iy – 4| – 16

|(x – 4) + iy| = 16

(x – 4)² + y² = 16²

∴ It is a circle with centre at (4, 0) and radius 16

x² + y² – 8x + 16 = 256

x² + y² – 8x – 240 = 0

(ii) |z – 4|² – |z – 1|² = 16.

Solution:

|x + iy – 4|² – |x + iy – 1|² = 16

|(x – 4) + iy|² – |(x – 1) + iy|² = 16

(x-4)² + y² – [(x – 1)² + y²] = 16

x² – 8x + 16 + y² – x² + 2x – 1 – y² = 16

-6x – 1 = 0 ⇒ 6x + 1 = 0