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

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

Question 1.
If z = 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:
$$\left|\frac{z-4 i}{z+4 i}\right|=1$$
⇒ |z – 4i| = |z + 4i|
let z = x + iy
⇒ |x + iy – 4i| = |x + iy + 4i|
⇒ |x + i(y – 4)| = |x +(y + 4)|
⇒ $$\sqrt{x^{2}+(y-4)^{2}}=\sqrt{x^{2}+(y+4)^{2}}$$
Squaring on both sides, we get
x2 + y2 – 8y + 16 = x2 + y2 + 16 + 8y
⇒ -16y = 0
⇒ y = 0 in two equation of real axis.

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

2x2 + 2y2 + x – 2y = 0.
Hence proved.

Question 3.
Obtain the Cartesian form of the locus of z = x + iy in each of the following cases:
(i) [Re(iz)]2 = 3
(ii) Im[(1 – i)z + 1] = 0
(iii) |z + i| = |z – 1|
(iv) $$\bar{z}=z^{-1}$$
Solution:
(i) Given z = x + iy
iz = i(x + iy)
∴ [Re (iz)]² = -y
(y)² = 3
y² = 3

(ii) Im[(1 – i)z + 1] = 0
Solution:
Given z = x + iy
= (1 – i) (x + iy) + 1
= x – ix + iy + y + 1 = 0
⇒ (x + y + 1) + i(y – x)
Im [(1 – i)z + 1] = 0
∴ y – x = 0
x – y = 0

(iii) |z + i| = |z – 1|
Solution:
|x + iy + i| = |x + iy – 1| [∵ z = x + iy]
|x + i(y + 1)| = |(x – 1) + iy|
$$\sqrt {x² + (y + 1)²}$$ = $$\sqrt {(x – 1)² + y²}$$
squaring on both sides
x² + (y + 1)² = (x – 1)² + y²
x² + y² + 2y + 1 = x² – 2x + 1 + y²
2x + 2y = 0
x + y = 0 [∵ z = x + iy]

(iv) $$\bar{z}=z^{-1}$$
⇒ $$\bar{z}=\frac{1}{z}$$
⇒ $$z \bar{z}=1$$
⇒ |z|2 = 1
⇒ |x + iy|2 = 1
⇒ x2 + y2 = 1

Question 4.
Show that the following equations represent a circle, and, find its centre and radius.
(i) |z – 2 – i| = 3
(ii) |2z + 2 – 4i| = 2
(iii) |3z – 6 + 12i| = 8
Solution:
(i) |z – 2 – i| = 3
This can be written as
|z – (2 – i)| = 3
This is the from |z – z0| = r and it represents a circle.
∴ Centre is (2, 1) and radius = 3 units
Aliter:
Let z = x + iy
∴ |z – 2 – i| = 3
|x + iy – 2 – i| = 3
|(x – 2) + i(y – 1)| = 3
$$\sqrt{(x+2)²+(y-1)²}$$ = 3
squaring on both sides
(x – 2)² + (y – 1)² = 9
x² – 4x + 4 + y² – 2y + 1 – 9 = 0
x² + y² – 4x – 2y – 4 = 0
Comparing with general from of equation of circle
ax² + by² + 2gx + 2fy + c = 0
we get a = 1, b = 1, g = -2, f = -1, c = -4
Centre (-g, -f) = (2, 1)
radius = $$\sqrt{g²+ f² – c}$$ = $$\sqrt{4 + 1 + 4}$$ = √9 = 3 units

(ii) |2z + 2 – 4i| = 2
Solution:
|2z + 2 – 4i| = 2
2 |z – (-1 + 2i)| = 2
This can be written as
|z – (-1 + 2i)| = 1
which is in the from |z – z0| = r
Centre of the circle = (-1, 2i)

(iii) |3z – 6 + 12i| = 8.
Solution:
3 |z – (2 + 4i)| = 8
⇒ |z – (2 – 4i)| = $$\frac{8}{3}$$
This is in the form |z – z0| = r
Centre of the circle (2 – 4i)
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
(ii) |z – 4|2 – |z – 1|2 = 16
Solution:
(i) z = x + iy
|z – 4| = 16
⇒ |x + iy – 4| = 16
⇒ |(x – 4) + iy| = 16
⇒ $$\sqrt{(x-4)^{2}+y^{2}}=16$$
Squaring on both sides
(x – 4)2 + y2 = 256
⇒ x2 – 8x + 16 + y2 – 256 = 0
⇒ x2 + y2 – 8x – 240 = 0 represents the equation of circle

(ii) Given z = x + iy
|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

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

Question 1.
If the imaginary part of is -2, then show that the locus of the point representing z in the argand plane is a straight line.
Solution:
Let z = x + iy. Then,

Hence, the locus of z is a straight line

Question 2.
If the real part of is 4, then show that the locus of the point representing z in the complex plane is a circle.
Solution:

It is given that the real part of is 4.

Question 3.

Solution:

Question 4.
If arg (z – 1) = $$\frac{\pi}{6}$$ and arg (z + 1) = 2 $$\frac{\pi}{3}$$ , then prove that |z| = 1.
Solution:

Question 5.
P represents the variable complex number z. Find the locus of P, if
Solution:

Question 6.
P represents the variable complex number z. Find the locus of P, if
Solution:
Let z = x + iy