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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) z = x + iy

[Re(iz)]^{2} = 3

⇒ [Re[i(x + iy]]^{2} = 3

⇒ [Re(ix – y)]^{2} = 3

⇒ (-y)^{2} = 3

⇒ y^{2} = 3

(ii) Im[(1 – i)z + 1] = 0

⇒ Im [(1 – i)(z + iy) + 1] = 0

⇒ Im[x + iy – ix + y + 1] = 0

⇒ Im[(x + y + 1) + i(y – x)] = 0

Considering only the imaginary part

y – x = 0 ⇒ x = y

(iii) |z + i| = |z – 1|

⇒ |x + iy + i| = | x + iy – 1|

⇒ |x + i(y + 1)| = |(x – 1) + iy|

Squaring on both sides

|x + i(y + 1)|^{2} = |(x – 1) + iy|^{2}

⇒ x^{2} + (y + 1)^{2} = (x – 1)^{2} + y^{2}

⇒ x^{2} + y^{2} + 2y + 1 = x^{2} – 2x + 1 + y^{2}

⇒ 2y + 2x = 0

⇒ x + y = 0

(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) 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

⇒ x^{2} – 4x + 4 + y^{2} – 2y + 1 – 9 = 0

⇒ x^{2} + y^{2} – 4x – 2y – 4 = 0 represents a circle

2g = -4 ⇒ g = -2

2f = -2 ⇒ f = -1

c = -4

(a) Centre (-g, -f) = (2, 1) = 2 + i

(b) Radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{4+1+4}=3\)

Aliter: |z – (2 + i)| = 3

Centre = 2 + i

radius = 3

(ii) |2(x + iy) + 2 – 4i| = 2

⇒ |2x + i2y + 2 – 4i| =2

⇒ |(2x + 2) + i(2y – 4)| = 2

⇒ |2(x + 1) + 2i(y – 2)| = 2

⇒ |(x + 1) + i(y – 2)| = 1

⇒ \(\sqrt{(x+1)^{2}(y-2)^{2}}=1\)

Squaring on both sides,

x^{2} + 2x + 1 + y^{2} + 4 – 4y – 1 = 0

⇒ x^{2} + y^{2} + 2x – 4y + 4 = 0 represents a circle

2g = 2 ⇒ g = 1

2f = -4 ⇒ f = -2

c = 4

(a) Centre (-g, -f) = (-1, 2) = -1 + 2i

(b) Radius = \(\sqrt{g^{2}+f^{2}-c}=\sqrt{1+4-4}=1\)

Aliter: 2|(z + 1 – 2i)| = 2

|z – (-1 + 2i)| = 1

Centre = -1 + 2i

radius = 1

(iii) |3(x + iy) – 6 + 12i| = 8

⇒ |3x + i3y – 6 + 12i| = 8

⇒ |3(x – 2) + i3 (y + 4)| = 8

⇒ 3|(x – 2) + i (y + 4)| = 8

⇒ \(3 \sqrt{(x-2)^{2}+(y+4)^{2}}=8\)

Squaring on both sides,

9[(x – 2)^{2} + (y + 4)^{2}] = 64

⇒ x^{2} – 4x + 4 + y^{2} + 8y + 16 = \(\frac{64}{9}\)

⇒ x^{2} + y^{2} – 4x + 8y + 20 – \(\frac{64}{9}\) = 0

x^{2} + y^{2} – 4x + 8y + \(\frac{116}{9}\) = 0 represents a circle.

2g = -4 ⇒ g = -2

2f = 8 ⇒ f = 4

c = \(\frac{116}{9}\)

(a) Centre (-g, -f) = (2, -4) = 2 – 4i

(b) Radius = \(=\sqrt{g^{2}+f^{2}-c}=\sqrt{4+16-\frac{116}{9}}=\sqrt{\frac{180-116}{9}}=\frac{8}{3}\)

Aliter:

|z – 2 + 4i| = \(\frac{8}{3}\)

⇒ |z – (2 – 4i)| = \(\frac{8}{3}\)

Centre = 2 – 4i, Radius = \(\frac{8}{3}\)

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) |x + iy – 4|^{2} – |x + iy – 1|^{2} = 16

⇒ |(x – 4) + iy|^{2} – |(x – 1) + iy|^{2} = 16

⇒ [(x – 4)^{2} + y^{2}] – [(x – 1)^{2} + y^{2}] = 16

⇒ (x^{2} – 8x + 16 + y^{2}) – (x^{2} – 2x + 1 + y^{2}) = 16

⇒ x^{2} + y^{2} – 8x + 16 – x^{2} + 2x – 1 – y^{2} = 16

⇒ -6x + 15 = 16

⇒ 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 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