RD Chapter 13 Complex Numbers Ex 13.2 Solutions
Question - 1 : - Express the following complex numbers in the standard form a + ib:
(i) (1 + i) (1 + 2i)
(ii) (3 + 2i) / (-2 + i)
(iii) 1/(2 + i)2
(iv) (1 – i) / (1 + i)
(v) (2 + i)3 / (2 + 3i)
(vi) [(1 + i) (1 +√3i)] / (1 – i)
(vii) (2 + 3i) / (4 + 5i)
(viii) (1 – i)3 / (1 – i3)
(ix) (1 + 2i)-3
(x) (3 – 4i) / [(4 – 2i) (1 + i)]
(xi)
(xii) (5 +√2i) / (1-√2i)
Answer - 1 : -
(i) (1 + i) (1 + 2i)
Let us simplify andexpress in the standard form of (a + ib),
(1 + i) (1 + 2i) =(1+i)(1+2i)
= 1(1+2i)+i(1+2i)
= 1+2i+i+2i2
= 1+3i+2(-1) [since, i2 =-1]
= 1+3i-2
= -1+3i
∴ The values of a,b are -1, 3.
(ii) (3 + 2i) / (-2 + i)
Let us simplify andexpress in the standard form of (a + ib),
(3 + 2i) / (-2 + i) =[(3 + 2i) / (-2 + i)] × (-2-i) / (-2-i) [multiply and divide with (-2-i)]
= [3(-2-i) + 2i(-2-i)] / [(-2)2 – (i)2]
= [-6 -3i – 4i -2i2]/ (4-i2)
= [-6 -7i -2(-1)] / (4– (-1)) [since, i2 = -1]
= [-4 -7i] / 5
∴ The values of a,b are -4/5, -7/5
(iii) 1/(2 + i)2
Let us simplify andexpress in the standard form of (a + ib),
1/(2 + i)2 =1/(22 + i2 + 2(2) (i))
= 1/ (4 – 1 + 4i)[since, i2 = -1]
= 1/(3 + 4i) [multiplyand divide with (3 – 4i)]
= 1/(3 + 4i) × (3 –4i)/ (3 – 4i)]
= (3-4i)/ (32 –(4i)2)
= (3-4i)/ (9 – 16i2)
= (3-4i)/ (9 – 16(-1))[since, i2 = -1]
= (3-4i)/25
∴ The values of a,b are 3/25, -4/25
(iv) (1 – i) / (1 + i)
Let us simplify andexpress in the standard form of (a + ib),
(1 – i) / (1 + i) = (1– i) / (1 + i) × (1-i)/(1-i) [multiply and divide with (1-i)]
= (12 +i2 – 2(1)(i)) / (12 – i2)
= (1 + (-1) -2i) / (1– (-1))
= -2i/2
= -i
∴ The values of a,b are 0, -1
(v) (2 + i)3 /(2 + 3i)
Let us simplify andexpress in the standard form of (a + ib),
(2 + i)3 /(2 + 3i) = (23 + i3 + 3(2)2(i) +3(i)2(2)) / (2 + 3i)
= (8 + (i2.i)+ 3(4)(i) + 6i2) / (2 + 3i)
= (8 + (-1)i + 12i +6(-1)) / (2 + 3i)
= (2 + 11i) / (2 + 3i)
[multiplyand divide with (2-3i)]
= (2 + 11i)/(2 + 3i) ×(2-3i)/(2-3i)
= [2(2-3i) +11i(2-3i)] / (22 – (3i)2)
= (4 – 6i + 22i – 33i2)/ (4 – 9i2)
= (4 + 16i – 33(-1)) /(4 – 9(-1)) [since, i2 = -1]
= (37 + 16i) / 13
∴ The values of a,b are 37/13, 16/13
(vi) [(1 + i) (1 +√3i)] /(1 – i)
Let us simplify andexpress in the standard form of (a + ib),
[(1 +i) (1 +√3i)] / (1 – i) = [1(1+√3i) + i(1+√3i)] / (1-i)
= (1 + √3i + i + √3i2)/ (1 – i)
= (1 + (√3+1)i +√3(-1)) / (1-i) [since, i2 = -1]
= [(1-√3) + (1+√3)i] /(1-i)
[multiplyand divide with (1+i)]
= [(1-√3) + (1+√3)i] /(1-i) × (1+i)/(1+i)
= [(1-√3) (1+i) +(1+√3)i(1+i)] / (12 – i2)
= [1-√3+ (1-√3)i +(1+√3)i + (1+√3)i2] / (1-(-1)) [since, i2 = -1]
= [(1-√3)+(1-√3+1+√3)i+(1+√3)(-1)]/ 2
= (-2√3 + 2i) / 2
= -√3 + i
∴ The values of a,b are -√3, 1
(vii) (2 + 3i) / (4 + 5i)
Let us simplify andexpress in the standard form of (a + ib),
(2 + 3i) / (4 + 5i) =[multiply and divide with (4-5i)]
= (2 + 3i) / (4 + 5i)× (4-5i)/(4-5i)
= [2(4-5i) + 3i(4-5i)]/ (42 – (5i)2)
= [8 – 10i + 12i – 15i2]/ (16 – 25i2)
= [8+2i-15(-1)] / (16– 25(-1)) [since, i2 = -1]
= (23 + 2i) / 41
∴ The values of a,b are 23/41, 2/41
(viii) (1 – i)3 /(1 – i3)
Let us simplify andexpress in the standard form of (a + ib),
(1 – i)3 /(1 – i3) = [13 – 3(1)2i + 3(1)(i)2 –i3] / (1-i2.i)
= [1 – 3i + 3(-1)-i2.i]/ (1 – (-1)i) [since, i2 = -1]
= [-2 – 3i – (-1)i] /(1+i)
= [-2-4i] / (1+i)
[Multiplyand divide with (1-i)]
= [-2-4i] / (1+i) ×(1-i)/(1-i)
= [-2(1-i)-4i(1-i)] /(12 – i2)
= [-2+2i-4i+4i2]/ (1 – (-1))
= [-2-2i+4(-1)] /2
= (-6-2i)/2
= -3 – i
∴ The values of a,b are -3, -1
(ix) (1 + 2i)-3
Let us simplify andexpress in the standard form of (a + ib),
(1 + 2i)-3 =1/(1 + 2i)3
= 1/(13+3(1)2 (2i)+2(1)(2i)2 +(2i)3)
= 1/(1+6i+4i2+8i3)
= 1/(1+6i+4(-1)+8i2.i)[since, i2 = -1]
= 1/(-3+6i+8(-1)i)[since, i2 = -1]
= 1/(-3-2i)
= -1/(3+2i)
[Multiplyand divide with (3-2i)]
= -1/(3+2i) ×(3-2i)/(3-2i)
= (-3+2i)/(32 –(2i)2)
= (-3+2i) / (9-4i2)
= (-3+2i) / (9-4(-1))
= (-3+2i) /13
∴ The values of a,b are -3/13, 2/13
(x) (3 – 4i) / [(4 – 2i)(1 + i)]
Let us simplify andexpress in the standard form of (a + ib),
(3 – 4i) / [(4 – 2i)(1 + i)] = (3-4i)/ [4(1+i)-2i(1+i)]
= (3-4i)/ [4+4i-2i-2i2]
= (3-4i)/ [4+2i-2(-1)][since, i2 = -1]
= (3-4i)/ (6+2i)
[Multiplyand divide with (6-2i)]
= (3-4i)/ (6+2i) ×(6-2i)/(6-2i)
= [3(6-2i)-4i(6-2i)] /(62 – (2i)2)
= [18 – 6i – 24i + 8i2]/ (36 – 4i2)
= [18 – 30i + 8 (-1)]/ (36 – 4 (-1)) [since, i2 = -1]
= [10-30i] / 40
= (1 – 3i) / 4
∴ The values of a,b are 1/4, -3/4
(xi) 
(xii) (5 +√2i) / (1-√2i)
Let us simplify andexpress in the standard form of (a + ib),
(5 +√2i) / (1-√2i) =[Multiply and divide with (1+√2i)]
= (5 +√2i) / (1-√2i) ×(1+√2i)/(1+√2i)
= [5(1+√2i) + √2i(1+√2i)] / (12 –(√2)2)
= [5+5√2i + √2i + 2i2] / (1 – 2i2)
= [5 + 6√2i + 2(-1)] / (1-2(-1)) [since, i2 =-1]
= [3+6√2i]/3
= 1+ 2√2i
∴ The values of a,b are 1, 2√2
Question - 2 : - Find the real values of x and y, if
(i) (x + iy) (2 – 3i) = 4 + i
(ii) (3x – 2i y) (2 + i)2 = 10(1 + i)
(iv) (1 + i) (x + iy) = 2 – 5i
Answer - 2 : -
(i) (x + iy) (2 – 3i) = 4+ i
Given:
(x + iy) (2 – 3i) = 4+ i
Let us simplify theexpression we get,
x(2 – 3i) + iy(2 – 3i)= 4 + i
2x – 3xi + 2yi – 3yi2 =4 + i
2x + (-3x+2y)i – 3y(-1) = 4 + i [since, i2 = -1]
2x + (-3x+2y)i + 3y =4 + i [since, i2 = -1]
(2x+3y) + i(-3x+2y) =4 + i
Equating Real andImaginary parts on both sides, we get
2x+3y = 4… (i)
And -3x+2y = 1… (ii)
Multiply (i) by 3 and(ii) by 2 and add
On solving we get,
6x – 6x – 9y + 4y = 12+ 2
13y = 14
y = 14/13
Substitute the valueof y in (i) we get,
2x+3y = 4
2x + 3(14/13) = 4
2x = 4 – (42/13)
= (52-42)/13
2x = 10/13
x = 5/13
x = 5/13, y = 14/13
∴ The real valuesof x and y are 5/13, 14/13
(ii) (3x – 2i y) (2 + i)2 =10(1 + i)
Given:
(3x – 2iy) (2+i)2 =10(1+i)
(3x – 2yi) (22+i2+2(2)(i))= 10+10i
(3x – 2yi) (4 +(-1)+4i) = 10+10i [since, i2 = -1]
(3x – 2yi) (3+4i) =10+10i
Let us divide with3+4i on both sides we get,
(3x – 2yi) =(10+10i)/(3+4i)
= Now multiply anddivide with (3-4i)
= [10(3-4i) +10i(3-4i)] / (32 – (4i)2)
= [30-40i+30i-40i2]/ (9 – 16i2)
= [30-10i-40(-1)] /(9-16(-1))
= [70-10i]/25
Now, equating Real andImaginary parts on both sides we get
3x = 70/25 and -2y =-10/25
x = 70/75 and y = 1/5
x = 14/15 and y = 1/5
∴ The real valuesof x and y are 14/15, 1/5
(4+2i) x-3i-3 +(9-7i)y = 10i
(4x+9y-3) + i(2x-7y-3)= 10i
Now, equating Real andImaginary parts on both sides we get,
4x+9y-3 = 0 … (i)
And 2x-7y-3 = 10
2x-7y = 13 … (ii)
Multiply (i) by 7 and(ii) by 9 and add
On solving theseequations we get
28x + 18x + 63y – 63y= 117 + 21
46x = 117 + 21
46x = 138
x = 138/46
= 3
Substitute the valueof x in (i) we get,
4x+9y-3 = 0
9y = -9
y = -9/9
= -1
x = 3 and y = -1
∴ The real valuesof x and y are 3 and -1
(iv) (1 + i) (x + iy) = 2 –5i
Given:
(1 + i) (x + iy) = 2 –5i
Divide with (1+i) onboth the sides we get,
(x + iy) = (2 –5i)/(1+i)
Multiply and divide by(1-i)
= (2 – 5i)/(1+i) ×(1-i)/(1-i)
= [2(1-i) – 5i (1-i)]/ (12 – i2)
= [2 – 7i + 5(-1)] / 2[since, i2 = -1]
= (-3-7i)/2
Now, equating Real andImaginary parts on both sides we get
x = -3/2 and y = -7/2
∴ Thee real valuesof x and y are -3/2, -7/2
Question - 3 : - Find the conjugates of the following complex numbers:
(i) 4 – 5i
(ii) 1 / (3 + 5i)
(iii) 1 / (1 + i)
(iv) (3 – i)2 / (2 + i)
(v) [(1 + i) (2 + i)] / (3 + i)
(vi) [(3 – 2i) (2 + 3i)] / [(1 + 2i) (2 – i)]
Answer - 3 : -
(i) 4 – 5i
Given:
4 – 5i
We know the conjugateof a complex number (a + ib) is (a – ib)
So,
∴ The conjugate of (4 –5i) is (4 + 5i)
(ii) 1 / (3 + 5i)
Given:
1 / (3 + 5i)
Since the givencomplex number is not in the standard form of (a + ib)
Let us convert tostandard form by multiplying and dividing with (3 – 5i)
We get,
We know the conjugateof a complex number (a + ib) is (a – ib)
So,
∴ The conjugate of (3 –5i)/34 is (3 + 5i)/34
(iii) 1 / (1 + i)
Given:
1 / (1 + i)
Since the givencomplex number is not in the standard form of (a + ib)
Let us convert tostandard form by multiplying and dividing with (1 – i)
We get,
We know the conjugateof a complex number (a + ib) is (a – ib)
So,
∴ The conjugate of(1-i)/2 is (1+i)/2
(iv) (3 – i)2 /(2 + i)
Given:
(3 – i)2 /(2 + i)
Since the givencomplex number is not in the standard form of (a + ib)
Let us convert tostandard form,
We know the conjugateof a complex number (a + ib) is (a – ib)
So,
∴ The conjugate of (2 –4i) is (2 + 4i)
(v) [(1 + i) (2 + i)] / (3+ i)
Given:
[(1 +i) (2 + i)] / (3 + i)
Since the givencomplex number is not in the standard form of (a + ib)
Let us convert tostandard form,
We know the conjugateof a complex number (a + ib) is (a – ib)
So,
∴ The conjugate of (3 +4i)/5 is (3 – 4i)/5
(vi) [(3 – 2i) (2 + 3i)] /[(1 + 2i) (2 – i)]
Given:
[(3 –2i) (2 + 3i)] / [(1 + 2i) (2 – i)]
Since the givencomplex number is not in the standard form of (a + ib)
Let us convert tostandard form,
We know the conjugateof a complex number (a + ib) is (a – ib)
So,
∴ The conjugate of (63– 16i)/25 is (63 + 16i)/25
Question - 4 : - Find the multiplicative inverse of the following complex numbers:
(i) 1 – i
(ii) (1 + i √3)2
(iii) 4 – 3i
(iv) √5 + 3i
Answer - 4 : -
(i) 1 – i
Given:
1 – i
We know themultiplicative inverse of a complex number (Z) is Z-1 or 1/Z
So,
∴ The multiplicativeinverse of (1 – i) is (1 + i)/2
(ii) (1 + i √3)2
Given:
(1 + i √3)2
Z = (1 + i √3)2
= 12 +(i √3)2 + 2 (1) (i√3)
= 1 + 3i2 +2 i√3
= 1 + 3(-1) + 2 i√3[since, i2 = -1]
= 1 – 3 + 2 i√3
= -2 + 2 i√3
We know themultiplicative inverse of a complex number (Z) is Z-1 or 1/Z
So,
Z = -2 + 2 i√3
∴ The multiplicativeinverse of (1 + i√3)2 is (-1-i√3)/8
(iii) 4 – 3i
Given:
4 – 3i
We know themultiplicative inverse of a complex number (Z) is Z-1 or 1/Z
So,
Z = 4 – 3i
∴ The multiplicativeinverse of (4 – 3i) is (4 + 3i)/25
(iv) √5 + 3i
Given:
√5 + 3i
We know themultiplicative inverse of a complex number (Z) is Z-1 or 1/Z
So,
Z = √5 + 3i
∴ The multiplicativeinverse of (√5 + 3i) is (√5 – 3i)/14
Question - 5 : - If z1 = (2 – i), z2 = (-2 + i), find
Answer - 5 : -
Given:
z1 =(2 – i) and z2 = (-2 + i)
Question - 6 : - Find the modulus of [(1 + i)/(1 – i)] – [(1 – i)/(1 + i)]
Answer - 6 : -
Given:
[(1 +i)/(1 – i)] – [(1 – i)/(1 + i)]
So,
Z = [(1 + i)/(1 – i)]– [(1 – i)/(1 + i)]
Let us simplify, weget
= [(1+i) (1+i) – (1-i)(1-i)] / (12 – i2)
= [12 +i2 + 2(1)(i) – (12 + i2 –2(1)(i))] / (1 – (-1)) [Since, i2 = -1]
= 4i/2
= 2i
We know that for acomplex number Z = (a+ib) it’s magnitude is given by |z| = √(a2 + b2)
So,
|Z| = √(02 + 22)
= 2
∴ The modulus of [(1 +i)/(1 – i)] – [(1 – i)/(1 + i)] is 2.
Question - 7 : - If x + iy = (a+ib)/(a-ib), prove that x2 + y2 =1
Answer - 7 : -
Given:
x + iy = (a+ib)/(a-ib)
We know that for acomplex number Z = (a+ib) it’s magnitude is given by |z| = √(a2 + b2)
So,
|a/b| is |a| / |b|
Applying Modulus onboth sides we get,
Question - 8 : - Find the least positive integral value of n for which [(1+i)/(1-i)]n isreal.
Answer - 8 : -
Given:
[(1+i)/(1-i)]n
Z = [(1+i)/(1-i)]n
Now let us multiplyand divide by (1+i), we get
= i [which is notreal]
For n = 2, we have
[(1+i)/(1-i)]2 =i2
= -1 [which is real]
So, the smallestpositive integral ‘n’ that can make [(1+i)/(1-i)]n real is2.
∴ The smallestpositive integral value of ‘n’ is 2.
Question - 9 : - Find the real values of θ for which the complex number (1 + i cos θ) / (1– 2i cos θ) is purely real.
Answer - 9 : -
Given:
(1 + i cos θ) / (1 –2i cos θ)
Z = (1 + i cos θ) / (1– 2i cos θ)
Let us multiply anddivide by (1 + 2i cos θ)
For a complex numberto be purely real, the imaginary part should be equal to zero.
So,
3cos θ = 0(since, 1 + 4cos2θ ≥ 1)
cos θ = 0
cos θ = cos π/2
θ = [(2n+1)π] / 2, forn ∈ Z
= 2nπ ± π/2, for n ∈ Z
∴ The values of θto get the complex number to be purely real is 2nπ ± π/2, for n ∈Z
Question - 10 : - Find the smallest positive integer value of n for which (1+i) n /(1-i) n-2 is a real number.
Answer - 10 : -
Given:
(1+i) n /(1-i) n-2
Z = (1+i) n /(1-i) n-2
Let us multiply anddivide by (1 – i)2
For n = 1,
Z = -2i1+1
= -2i2
= 2, which is a realnumber.
∴ The smallestpositive integer value of n is 1.