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## Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.4

Question 1.

Write the following in the rectangular form:

(i) \(\overline{(5+9 i)+(2-4 i)}\)

(ii) \(\frac{10-5 i}{6+2 i}\)

(iii) \(\overline{3 i}+\frac{1}{2-i}\)

Solution:

Question 2.

If z = x + iy, find the following in rectangular form.

(i) Re(\(\frac{1}{z}\))

(ii) Re(i\(\bar{z}\))

(iii) Im(3z + 4\(\bar{z}\) – 4i)

Solution:

(i) Re(\(\frac{1}{z}\)) = Re(\(\frac{1}{x+i y} \times \frac{x-i y}{x-i y}\))

= Re(\(\frac{x-i y}{x^{2}+y^{2}}\))

= \(\frac{x}{x^{2}+y^{2}}\)

(ii) Re(i\(\bar{z}\)) = Re[i(\(\overline{x+i y}\))]

= Re(ix + y)

= y

(iii) Im(3z + 4\(\bar{z}\) – 4i)

= Im (3(x + iy) + 4(x – iy) – 4i)

= Im (3x + 3iy + 4x – 4iy – 4i)

= Im (3x + 4 + i (3y – 4y – 4)

= Im (3x + 4x + i(-y – 4))

= Im [7x + i(-y – 4)]

= -y – 4

= -(y + 4)

Question 3.

If z_{1} = 2 – i and z_{2} = -4 + 3i, find the inverse of z_{1} z_{2} and \(\frac{z_{1}}{z_{2}}\)

Solution:

z_{1} = 2 – i, z_{2} = -4 + 3i

(i) z_{1} z_{2} = (2 – i) (-4 + 3i)

= (-8 + 6i + 4i – 3 i^{2})

= (-8 + 10i + 3)

= (-5 + 10i)

Question 4.

The complex numbers u, v, and w are related by \(\frac{1}{u}=\frac{1}{v}+\frac{1}{w}\). If v = 3 – 4i and w = 4 + 3i, find u in rectangular form.

Solution:

v = 3 – 4i, w = 4 + 3i = i (3 – 4i)

Question 5.

Prove the following properties:

(i) z is real if and only if z = \(\bar{z}\)

(ii) Re(z) = \(\frac{z+\bar{z}}{2}\) and Im(z) = \(\frac{z-\bar{z}}{2 i}\)

Solution:

(i) z is real iff z = \(\bar{z}\)

Let z = x + iy

z = \(\bar{z}\)

⇒ x + iy = x – iy

⇒ 2iy = 0

⇒ y = 0

⇒ z is real.

z is real iff z = \(\bar{z}\)

(ii) \(\frac{z+\bar{z}}{2 i}=\frac{x+i y+x-i y}{2}=\frac{2 x}{2}=x\)

Real part of z = x

(iii) \(\frac{z-\bar{z}}{2 i}=\frac{(x+i y)-(x-i y)}{2 i}=\frac{x+i y-x+i y}{2 i}=\frac{2 i y}{2 i}=y\)

Im part of z = y.

Question 6.

Find the least value of the positive integer n for which (√3 + i)^{n}

(i) real

(ii) purely imaginary

Solution:

(√3 + i)^{n}

(√3 + i)^{2}

= 3 – 1 + 2√3 i

= (2 + 2 √3 i)

(√3 + i)^{3} = (√3 + i)^{2} (√3 + i)

= (2 + 2√3 i) (√3 + i)

= 2√3 + 2i + 6i – 2√3

(√3 + i) = 8i ⇒ purely Imaginary when n = 3

(√3 + i)^{4} = (√3 + i)^{3} (√3 + i)

= 8i (√3 + i)

= (-8 + 8√3 i)

(√3 + i)^{5} =(√3 + i)^{4} (√3 + i)

= (-8 + 8√3 i) (√3 + i)

= -8√3 – 8i + 24i – 8√3

= -16√3 + 16i

(√3 + i)^{6} = (√3 + i)^{5 }(√3 + i)

= (√3 + i) (-16√3 + 16i)

= 16 (√3 + i) (-√3 + i)

= 16 (-3 + i√3 – i√3 – 1)

= -64 purely real when n = 6

Another Method:

Question 7.

Show that

(i) (2 + i√3)^{10} – (2 – i√3)^{10} is purely imaginary

(ii) \(\left(\frac{19-7 i}{9+i}\right)^{12}+\left(\frac{20-5 i}{7-6 i}\right)^{12}\)

Solution:

(i) (2 + i√3)^{10} – (2 – i√3)^{10}

### Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.4 Additional Problems

Question 1.

Express the following in the standard form a + ib.

Solution:

Question 2.

Find the least positive integer n such that

Solution:

Question 3.

Find the real values of x and y for which the following equations are satisfied.

Solution:

(i) (1 – i)x + (1 + i)y = x – ix + y + iy

= (x + y) + i (y – x) = 1 – 3i (given)

So, equating their RP and IP we get,

Take real part, we get

i.e., 3x + (x – 2) + 6y – (1 – 3y) = 0

⇒ 3x + x – 2 + 6y – 1 + 3y = 0

4x + 9y = 3 …….. (1)

Take imaginary part, we get

3(x – 2) – x + 3 (1 – 3y) + 2y = 10

⇒ 3x – 6 – x + 3 – 9y + 2y = 10

Squaring on both sides, x^{2} + 3x + 8 = 4 (x + 4)^{2}

i.e., x^{2} + 3x +8 = 4 (x^{2} + 8x +16) ⇒ 4x^{2} + 32x + 64 – x^{2} – 3x – 8 = 0

3x^{2} + 29x + 56 = 0

3x^{2} + 21x + 8x + 56 = 0

(x + 7) (3x + 8) = 0