# CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a)

Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Ex 6(a) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 11 Math Solutions Chapter 6 Complex Numbers and Quadratic Equations Exercise 6(a)

Question 1.
Multiply (2√-3 + 3√-2) by (4√-3 – 5√-2)
Solution:
(2√-3 + 3√-2) by (4√-3 – 5√-2)
= (2√3i + 3√2i) (4√3i – 5√2i)
= i2 (2√3 + 3√2) (4√3 – 5√2)
= – 1(24 – 10√6 + 12√6 – 30)
= – 1(- 6 + 2√6) = 6 – 2√6

Question 2.
Multiply (3√-7 – 5√-2) (3√-2 + 5√-2)
Solution:
(3√-7 – 5√-2) (3√-2 + 5√-2)
= (3√7i – 5√-2i) (3√2i + 5√2i)
= i2 (3√7- 5√2) (3√2 + 5√2)
= (- 1)8√2(3√7 – 5√2 )
= – 24√14 + 80

Question 3.
Multiply (√-1 +√-1) (a – b√-1)
Solution:
(√-1 +√-1) (a – b√-1)
= (i + i) (a – bi) = 2i(a – bi)
= 2ai – 2bi2 = 2ai + 2b

Question 4.
Multiply (x – $$\frac{1+\sqrt{-3}}{2}$$) (x – $$\frac{1-\sqrt{3}}{2}$$)
Solution:
(x – $$\frac{1+\sqrt{-3}}{2}$$) (x – $$\frac{1-\sqrt{3}}{2}$$)
= (x + $$\frac{-1-i \sqrt{3}}{2}$$) (x + $$\frac{-1+i \sqrt{3}}{2}$$)
= (x + ω) (x + ω2) = x2 + ω2x + ωx + ω3
= x2 + x (ω 2 + ω) + 1 = x2 –  x + 1

Question 5.
Express with rational denominator. $$\frac{1}{3-\sqrt{-2}}$$
Solution:
\begin{aligned} & \frac{1}{3-\sqrt{-2}}=\frac{1}{3-\sqrt{2} \mathrm{i}}=\frac{3+\sqrt{2} \mathrm{i}}{(3-\sqrt{2} \mathrm{i})(3+\sqrt{2} \mathrm{i})} \\ & =\frac{3+\sqrt{2} \mathrm{i}}{9-2 \mathrm{i}^2}=\frac{3+\sqrt{2} \mathrm{i}}{9+2}=\frac{3+\mathrm{i} \sqrt{2}}{11} \end{aligned}

Question 6.
$$\frac{3 \sqrt{-2}+2(-5)}{3 \sqrt{-2}-2 \sqrt{-2}}$$
Solution:

Question 7.
$$\frac{3+2 \sqrt{-1}}{2-5 \sqrt{-1}}+\frac{3-2 \sqrt{-1}}{2+5 \sqrt{-1}}$$
Solution:

Question 8.
$$\frac{a+x \sqrt{-1}}{a-x \sqrt{-1}}-\frac{a-x \sqrt{-1}}{a+x \sqrt{-1}}$$
Solution:

Question 9.
$$\frac{(x+\sqrt{-1})^2}{x-\sqrt{-1}}-\frac{\left(x-\sqrt{-1^2}\right)}{x+\sqrt{-1}}$$
Solution:

Question 10.
$$\frac{(a+\sqrt{-1})^3-(a-\sqrt{-1})^3}{(a+\sqrt{-1})^2-(a-\sqrt{-1})^2}$$
Solution:

Question 11.
Find the value of (- i)4n+3; when n is positive.
Solution:
(- i)4n+3
= (-i4n) (-i)3 = 1(- i3) = – (-i) = i

Question 12.
Find the square root of (a + 40i) + $$\sqrt{9-40 \sqrt{-i}}$$
Solution:

Question 13.
Express in the form of a + ib:
(i) $$\frac{3+5 i}{2-3 i}$$
Solution:

(ii) $$\frac{\sqrt{3}-i \sqrt{2}}{2 \sqrt{3}-i \sqrt{3}}$$
Solution

(iii) $$\frac{(\mathrm{I}+\mathrm{i})^2}{3-\mathrm{i}}$$
Solution:

(iv) $$\frac{(a+i b)^3}{a-i b}-\frac{(a-i b)^3}{a+i b}$$
Solution:

(v) $$\frac{1+i}{1-i}$$
Solution:
$$\frac{1+i}{1-i}$$ = $$\frac{(1+i)^2}{2}=\frac{1-1+2 i}{2}$$
= i = 0 + i

Question 14.
Express the following points geometrically in the Argand plane.
(i) 1
Solution:
1 = 1 + i0 = (1, 0)

(ii) 3i
Solution:
3i = 0 + 3i = (0, 3)

(iii) – 2
Solution:
– 2 = – 2 + i0 = (- 2, 0)

(iv) 3 + 2i
Solution:
3 + 2i = (3, 2)

(v) – 3 + i
Solution:
– 3 + i = (- 3, 1)

(vi) 1-i
Solution:
1 – i = (1, – 1)

Question 15.
Show that the following numbers are equidistant from the origin:
√2 +i, 1 + i√2, i√3
Solution:
|√2 + i| = $$\sqrt{(\sqrt{2})^2+1^2}$$ = √3
|1 + i√2| = $$\sqrt{1^2+(\sqrt{2})^2}$$ = √3
and |i√3| = √3
∴ The points are equidistant from the origin.

Question 16.
Express each of the above complex numbers in the polar form.
Solution:

Question 17.
If 1, ω, ω2 are the three cube roots of unity, prove that (1 + ω2)4 = ω
Solution:
L.H.S. = (1 + ω2)4 = (- ω)4 = ω4
= ω3 .ω = 1. ω = ω
= R.H.S. (Proved)

Question 18.
(1 – ω+ ω2 ) (1 + ω – ω2 ) = 4
Solution:
L.H.S. = (1 – ω+ ω2 ) (1 + ω – ω2 )
= (- ω – ω )(- ω2 – ω2 ) (∴ 1 + ω + ω2 = 0)
= (- 2ω2 – 2ω2) = 4ω3 = 4 = R.H.S.

Question 19.
(1 – ω) (1 – ω)2 (1 – ω4) (1 – ω5) = 9
Solution:
L.H.S. =
(1 – ω) (1 – ω)2 (1 – ω4) (1 – ω5)
= (1 – ω) (1 – ω2) (1 – ω) (1 – ω2)
= (1 – ω)2 (1 – ω2)2
= {(1 – ω) (1 – ω2)}2
= (1 – ω2 – ω + ω3)2
= {3 – (ω2 + ω + 1)}2
= (3)2 = 9 = R.H.S.

Question 20.
(2 + 5ω + 2ω2 )6 = (2 + 2ω + 5ω2 )6 =729
Solution:
L.H.S. = (2 + 5ω + 2ω2 )6
(2 + 2ω2 + 5ω)6 = {2(1 + ω2 ) + 5ω}6
(- 2ω + 5ω)6 = (3ω)6 = 729ω6 = 729
Again, (2 + 2ω + 5ω2 )6
= {2(1 + ω) + 5ω2 )6
= (- 2ω2 + 5ω2 )6 = (3ω2)6
= 729ω12 =729
∴ (2 + 5ω + 2ω2 )6 = (2 + 2ω + 5ω2 )6 =729

Question 21.
(1 – ω + ω2 ) (1 – ω2 + ω4) (1 – ω42)  ….to 2n factors = 22n
Solution:
L.H.S. = (1 – ω + ω2 ) (1 – ω2 + ω4) (1 – ω42)  ….to 2n factors = 22n
= (- ω – ω) (1 – ω2 + ω) (1 – ω + ω2 )  ….to 2n factors = 22n
= (- 2ω) (- ω2 – ω2 ) (- ω – ω)  ….to 2n factors = 22n
= [(- 2ω)(- 2ω) … to n factors] × [(- 2ω2)(- 2ω2) …. to n factors]
= (- 2ω)n × (- 2ω2)n = (4ω3)n = 4n = 22n
R.H.S. (Proved)

Question 22.
Prove that x3 + y3 + z3– 3xyz
= (x + y + z) (x + ωy + ω2z) (x + yω2 + zω)
Solution:
R.H.S. = (x + y + z) (x + ωy + ω2z) (x + yω2 + zω)
= (x + y + z) (x + xyω2 + zxω + xyω + y2ω3+ yz ω2 + zxω2 + yzω4 + z2ω3)
= (x + y + z) [x2 + y2 + z2 + xy (ω2 + ω) +yz (ω2 + ω) + zx(ω2 + ω)]
= (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
= x3 + y3 + z3 – 3xyz = L.H.S. (Proved)

Question 23.
If x = a + b, y = aω + b ω2, z = aω2 + bω show that
(1) xyz = a3 + b3
Solution:
L.H.S. = xyz
= (a + b) (a ω + b ω2) (a ω2 + b ω)
= (a + b) (a2 ω3 + ab ω2 + ab ω4 + b2 ω3)
= (a + b) {a2 + b2 + ab(ω2 + ω)}
= (a + b) (a2 – ab + b2) = a3 + b3 = R.H.S. (Proved)

(2) x2 + y2 + z2 = 6ab
Solution:
L.H.S. = x2 + y2 + z2
= (a + b)2 + (a ω + b ω2)2 (a ω2 + b ω)2
= a2 + b2 + 2ab + a2 ω2 + b2 ω4 + 2ab ω3 + a2 ω4 + b2 ω2 + 2ab ω3
= a2 + a2 ω2 + a2 ω + b2 + b2 ω + b2 ω2 + 2ab + 2ab + 2ab
= a2(1 + ω2 + ω) + b2 (1 + ω + ω2) + 6ab
= 0 + 0 + 6ab = 6ab = R.H.S. (Proved)

(3) x3 + y3 + z3 = 3(a3 + b3)
Solution:
L.H.S. = x3 + y3 + z3
= (a + b)3 + (a ω + b ω2)3 + (aω2 + b2)3
= a3 + 3a2b + 3ab2 + b3 + a3 ω3 + 3a2b22 + 3a ωb2 ω4 + b3 ω6 + a3 ω6 + 3a2 ω4bω + 3a ω2b2 ω2 + b3 ω3
= a3 + a3 + a3 + b3 + b3 + b3 + 3a2b(1 + ω4 + ω5) + 3ab2 (1 + ω5 + ω4)
= 3a3 + 3b3 + 0 + 0 = 3 (a3 + b3) = R.H.S. (Proved)

Question 24.
If ax + by + cz = X, cx + by + az = Y, bx + ay + cz = Z
show that (a2 + b2 + c2 – ab – bc- ca) (x2 +y2 + z2 – xy – yz – zx) = X2 + Y2 + Z2 – YZ – ZX – XY
Solution:
L.H.S.
= (a2 + b2 + c2 – ab – bc – ca) (x2 + y2 + z2 – xy – yz – zx)
= (a + b ω + cω2) (a + bω2 + cω) (x + yω + zω)2 (x + yω2 + z ω) (Refer Q.No.22)
= {(a + bω + cω2) (x + yω + zω2)} {(a + bω2 + cω) (x + yω2 + zω)}
= (ax + ayω + azω2 + bx ω + byω2 + bzω3 + cxω2 + cyω3 + czω4) × (ax + ay ω2 + azω + bxω2 + byω4 + bzω3 + cxω + cyω3 + czω2)
= {(ax + cy + bz) + (cx + by + az) ω + (bx + ay + cz) ω2} x (ax + cy + bz) + (cx + by + az) ω + (bx + ay + cz) ω2}
= (X + Yω2 + Zω) (X + Yω + Zω2)
= X2 + Y2 + Z2 – XY – YZ – ZX (Refer Q. No. 22)
(Where X = ax + cy + bz.
Y = cx + by + az.
Z = bx + ay + cz).