Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b)

Practicing the Intermediate 2nd Year Maths 2A Textbook Solutions Inter 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(b) will help students to clear their doubts quickly.

Intermediate 2nd Year Maths 2A Complex Numbers Solutions Exercise 1(b)

I.

Question 1.
Write the following complex numbers in the form A + iB.
(i) (2 – 3i) (3 + 4i)
(ii) (1 + 2i)3
(iii) \(\frac{a-i b}{a+i b}\)
(iv) \(\frac{4+3 i}{(2+3 i)(4-3 i)}\)
(v) (-√3 + √-2) (2√3 – i)
(vi) (-5i) \(\left(\frac{i}{8}\right)\)
(vii) (-i) 2i
(viii) i9
(ix) i-19
(x) 3(7 + 7i) + i(7 + 7i)
(xi) \(\frac{2+5 i}{3-2 i}+\frac{2-5 i}{3+2 i}\)
Solution:
(i) (2 – 3i) (3 + 4i) = 6 + 8i – 9i – 12i2
= 6 – i + 12
= 18 – i
= 18 + i(-1)

(ii) (1 + 2i)3 = 1 + 3 . i2 . 2i + 3 . 1 . 4i2 + 8i3
= 1 + 6i – 12 – 8i
= -11 – 2i
= (-11) + i(-2)

Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) I Q1

Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) I Q1.1

(v) (-√3 + √-2) (2√3 – i) = (-√3 + i√2) (2√3 – i)
= -6 + i√3 + i2√6 + √2
= (-6 + √2) + i(√3 + 2√6)

(vi) (-5i) \(\left(\frac{i}{8}\right)\) = \(\frac{-5 i^{2}}{8}\)
= \(\frac{5}{8}\) + i(0)

(vii) (-i)(2i) = -2i2
= -2(-1)
= 2
= 2 + i(0)

(viii) i9 = i4 . i4 . i
= 1 . 1 . i
= i
= 0 + i(1)

Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b)

(ix) i-19 = \(\frac{1}{i^{19}}\)
= \(\frac{i}{i^{20}}\)
= \(\frac{i}{\left(i^{4}\right)^{5}}\)
= \(\frac{\mathrm{i}}{\mathrm{i}^{5}}\)
= i
= 0 + i(1)

(x) 3(7 + 7i) + i(7 + 7i)
= 21 + 21i + 7i – 7
= 14 + 28i

Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) I Q1.2

Question 2.
Write the conjugate of the following complex numbers.
(i) 3 + 4i
(ii) (15 + 3i) – (4 – 20i)
(iii) (2 + 5i) (-4 + 6i)
(iv) \(\frac{5 i}{7+i}\)
Solution:
(i) Let z = 3 + 4i
\(\bar{z}\) = 3 – 4i

(ii) Let z = (15 + 3i) – (4 – 20i)
= 15 + 3i – 4 + 20i
= 11 + 23i
\(\bar{z}\) = 11 – 23i

(iii) Let z = (2 + 5i) (-4 + 6i)
= -8 + 12i – 20i – 30
= -38 – 8i
\(\bar{z}\) = -38 + 8i
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) I Q2

Question 3.
Simplify
(i) i2 + i4 + i6 + …….. + (2n + 1) terms
(ii) i18 – 3 . i7 + i2 (1 + i4) (-i)26
Solution:
(i) i2 + i4 + i6 + …….. + (2n + 1) terms
= -1 + 1 – 1 + (2n + 1) terms
= -1

(ii) i18 – 3i2 + i2 (1 + i4) (-i)26
= i16 . i2 – 3 . i4 . i3 + i2 (1 + 1) i24 . i2
= 1 . (-1) – 3 . 1 . (-i) + (-1) (2) (1) (-1)
= -1 + 3i + 2
= 1 + 3i

Question 4.
Find a square root for the following complex numbers.
(i) 7 + 24i
(ii) -8 – 6i
(iii) (3 + 4i)
(iv) (-47 + i8√3)
Solution:
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) I Q4
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) I Q4.1
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) I Q4.2

Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b)

Question 5.
Find the multiplicative inverse of the following complex numbers.
(i) √5 + 3i
(ii) -i
(iii) i-35
Solution:
(i) √5 + 3i
The multiplicative inverse of x + iy is \(\frac{x-i y}{x^{2}+y^{2}}\)
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) I Q5

II.

Question 1.
(i) If (a + ib)2 = x + iy, find x2 + y2
Solution:
(a + ib)2 = x + iy
⇒ a2 + i(2ab) – b2 = x + iy
⇒ (a2 – b2) + i(2ab) = x + iy
⇒ (a2 – b2) + i(2ab) = x + iy
Equating real and imaginary parts on both sides, we have
x = a2 – b2 and y = 2ab
x2 + y2 = (a2 – b2)2 + (2ab)2
= a4 – 2a2b2 + b4 + 4a2b2
= a4 + 2a2b2 + b4
= (a2 + b2)2

(ii) If x + iy = \(\frac{3}{2+\cos \theta+i \sin \theta}\) then show that x2 + y2 = 4x – 3
Solution:
x + iy = \(\frac{3}{2+\cos \theta+i \sin \theta}\) \(\frac{(2+\cos \theta)-i \sin \theta}{(2+\cos \theta)-i \sin \theta}\)
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) II Q1(ii)
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) II Q1(ii).1

(iii) If x + iy = \(\frac{1}{1+\cos \theta+i \sin \theta}\), show that 4x2 – 1 = 0.
Solution:
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) II Q1(iii)
Equating real parts on both sides, we have
x = \(\frac{1}{2}\)
⇒ 2x = 1
⇒ 4x2 = 1
⇒ 4x2 – 1= 0

(iv) If u + iv = \(\frac{2+i}{z+3}\) and z = x + iy, find u, v.
Solution:
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) II Q1(iv)
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) II Q1(iv).1

Question 2.
(i) If z = 3 – 5i Show that z3 – 10z2 + 58z – 136 = 0.
Solution:
Given z = 3 – 5i
⇒ z – 3 = -5i
⇒ (z – 3)2 = 25i2
⇒ z2 – 6z + 9 = -25
⇒ z2 – 6z + 34 = 0
∴ z3 – 10z2 + 58z – 136 = z(z2 – 6z + 34) – 4z2 + 24z – 136
= z(0) – 4(z2 – 6z + 34)
= 0 – 4(0)
= 0
∴ z3 – 10z2 + 58z – 136 = 0

Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b)

(ii) If z = 2 – i√7 , then show that 3z3 – 4z2 + z + 88 = 0.
Solution:
Given z = 2 – i√7
⇒ z – 2 = -i√7
⇒ (z – 2)2 = (-i√7)2
⇒ z2 – 4z + 4 = 7i2
⇒ z2 – 4z + 4 = -7
⇒ z2 – 4z + 11 = 0
∴ 3z3 – 4z2 + z + 88 = 3z(z2 – 4z + 11) + 8z2 – 32z + 88
= 3z(0) + 8(z2 – 4z + 11)
= 0 + 8(0)
= 0
∴ 3z3 – 4z2 + z + 88 =0

(iii) Show that \(\frac{2-i}{(1-2 i)^{2}}\) and \(\left(\frac{-2-11 i}{25}\right)\) conjugate to each other.
Solution:
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) II Q2(iii)

Question 3.
(i) If (x – iy)1/3 = a – ib then show that \(\frac{x}{a}+\frac{y}{b}\) = 4(a2 – b2)
Solution:
Given (x – iy)1/3 = (a – ib)
⇒ x – iy = (a – ib)3
⇒ x – iy = a3 – 3a2ib + 3ai2b2 – i3b3
⇒ x – iy = (a3 – 3ab2) – i(3a2b – b3)
Equating real and imaginary parts
x = a3 – 3ab2
⇒ \(\frac{x}{a}\) = a2 – 3b2
y = 3a2b – b3
⇒ \(\frac{y}{b}\) = 3a2 – b2
∴ \(\frac{x}{a}+\frac{y}{b}\) = a2 – 3b2 + 3a2 – b2
= 4a2 – 4b2
= 4(a2 – b2)
∴ \(\frac{x}{a}+\frac{y}{b}\) = 4(a2 – b2)

Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b)

(ii) Write \(\left(\frac{a+i b}{a-i b}\right)^{2}-\left(\frac{a-i b}{a+i b}\right)^{2}\) in the form of x + iy.
Solution:
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) II Q3(ii)

(iii) If x and y are real numbers such that \(\frac{(1+i) x-2 i}{3+i}+\frac{(2-3 i) y+i}{3-1}=i\), then determine the values of x and y.
Solution:
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) II Q3(iii)
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) II Q3(iii).1

Question 4.
(i) Find the least positive integer n, satisfying \(\left(\frac{1+i}{1-i}\right)^{n}\) = 1
Solution:
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) II Q4(i)

(ii) If \(\left(\frac{1+i}{1-i}\right)^{3}-\left(\frac{1-i}{1+i}\right)^{3}\) = x + iy, find x and y.
Solution:
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) II Q4(ii)
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) II Q4(ii).1

(iii) Find real values of ‘θ’ in order that \(\frac{3+2 i \sin \theta}{1-2 i \sin \theta}\) is a
(a) real numbers
(b) Purely imaginary number
Solution:
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) II Q4(iii)
Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b) II Q4(iii).1

Inter 2nd Year Maths 2A Complex Numbers Solutions Ex 1(b)

(iv) Find the real values of x and y if \(\frac{x-1}{3+i}+\frac{y-1}{3-i}=i\)
Solution:
Given \(\frac{x-1}{3+i}+\frac{y-1}{3-i}=i\)
⇒ \(\frac{(x-1)(3-i)+(y-1)(3+i)}{9-i^{2}}=i\)
⇒ 3x – xi – 3 + i + 3y – iy – 3 – i = 10i
⇒ (3x + 3y – 6) + i(-x + y) = 0 + 10i
Now equating real and imaginary parts
3x + 3y – 6 = 0
⇒ x + y – 2 = 0 ……..(1)
-x + y = 10
⇒ x – y + 10 = 0 ………(2)
(1) + (2) ⇒ 2x + 8 = 0
⇒ x = -4
From (1),
-4 + y – 2 = 0
⇒ y = 6
∴ x = -4, y = 6

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