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## 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

x^{2} + y^{2} – 8y + 16 = x^{2} + y^{2} + 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 2x^{2} + 2y^{2} + x – 2y = 0.

Solution:

Let z = x + iy

2x^{2} + 2y^{2} + 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

⇒ x^{2} + y^{2} = 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 – z_{0}| = 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 – z_{0}| = r

Centre of the circle = (-1, 2i)

radius = 1 unit

(iii) |3z – 6 + 12i| = 8.

Solution:

3 |z – (2 + 4i)| = 8

⇒ |z – (2 – 4i)| = \(\frac{8}{3}\)

This is in the form |z – z_{0}| = 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} + y^{2} = 256

⇒ x^{2} – 8x + 16 + y^{2} – 256 = 0

⇒ x^{2} + y^{2} – 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