RD Chapter 13 Complex Numbers Ex 13.4 Solutions
Question - 1 : - Find the modulus and arguments of the following complex numbers and henceexpress each of them in the polar form:
(i) 1 + i
(ii) √3 + i
(iii) 1 – i
(iv) (1 – i) / (1 + i)
(v) 1/(1 + i)
(vi) (1 + 2i) / (1 – 3i)
(vii) sin 120o – i cos 120o
(viii) -16 / (1 + i√3)
Answer - 1 : -
We know that the polarform of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ)
Where,
|Z| = modulus ofcomplex number = √(x2 +y2)
θ = arg (z) = argumentof complex number = tan-1 (|y| / |x|)
(i) 1 + i
Given: Z = 1 + i
So now,
|Z| = √(x2 + y2)
= √(12 + 12)
= √(1 + 1)
= √2
θ = tan-1 (|y|/ |x|)
= tan-1 (1/ 1)
= tan-1 1
Since x > 0, y >0 complex number lies in 1st quadrant and the value of θ is 00≤θ≤900.
θ = π/4
Z = √2 (cos (π/4) + i sin (π/4))
∴ Polar form of (1 + i)is √2 (cos (π/4) + i sin(π/4))
(ii) √3 + i
Given: Z = √3 + i
So now,
|Z| = √(x2 + y2)
= √((√3)2 + 12)
= √(3 + 1)
= √4
= 2
θ = tan-1 (|y|/ |x|)
= tan-1 (1/ √3)
Since x > 0, y >0 complex number lies in 1st quadrant and the value of θ is 00≤θ≤900.
θ = π/6
Z = 2 (cos (π/6) + isin (π/6))
∴ Polar form of (√3 +i) is 2 (cos (π/6) + i sin (π/6))
(iii) 1 – i
Given: Z = 1 – i
So now,
|Z| = √(x2 + y2)
= √(12 + (-1)2)
= √(1 + 1)
= √2
θ = tan-1 (|y|/ |x|)
= tan-1 (1/ 1)
= tan-1 1
Since x > 0, y <0 complex number lies in 4th quadrant and the value of θ is -900≤θ≤00.
θ = -π/4
Z = √2 (cos (-π/4) + i sin (-π/4))
= √2 (cos (π/4) – i sin (π/4))
∴ Polar form of (1 – i)is √2 (cos (π/4) – i sin(π/4))
(iv) (1 – i) / (1 + i)
Given: Z = (1 – i) /(1 + i)
Let us multiply anddivide by (1 – i), we get
= 0 – i
So now,
|Z| = √(x2 + y2)
= √(02 + (-1)2)
= √(0 + 1)
= √1
θ = tan-1 (|y|/ |x|)
= tan-1 (1/ 0)
= tan-1 ∞
Since x ≥ 0, y < 0complex number lies in 4th quadrant and the value of θ is -900≤θ≤00.
θ = -π/2
Z = 1 (cos (-π/2) + isin (-π/2))
= 1 (cos (π/2) – i sin(π/2))
∴ Polar form of (1 – i)/ (1 + i) is 1 (cos (π/2) – i sin (π/2))
(v) 1/(1 + i)
Given: Z = 1 / (1 + i)
Let us multiply anddivide by (1 – i), we get
So now,
|Z| = √(x2 + y2)
= √((1/2)2 + (-1/2)2)
= √(1/4 + 1/4)
= √(2/4)
= 1/√2
θ = tan-1 (|y|/ |x|)
= tan-1 ((1/2)/ (1/2))
= tan-1 1
Since x > 0, y <0 complex number lies in 4th quadrant and the value of θ is -900≤θ≤00.
θ = -π/4
Z = 1/√2 (cos (-π/4) + i sin (-π/4))
= 1/√2 (cos (π/4) – i sin (π/4))
∴ Polar form of 1/(1 +i) is 1/√2 (cos (π/4) – i sin(π/4))
(vi) (1 + 2i) / (1 – 3i)
Given: Z = (1 + 2i) /(1 – 3i)
Let us multiply anddivide by (1 + 3i), we get
So now,
|Z| = √(x2 + y2)
= √((-1/2)2 + (1/2)2)
= √(1/4 + 1/4)
= √(2/4)
= 1/√2
θ = tan-1 (|y|/ |x|)
= tan-1 ((1/2)/ (1/2))
= tan-1 1
Since x < 0, y >0 complex number lies in 2nd quadrant and the value of θ is 900≤θ≤1800.
θ = 3π/4
Z = 1/√2 (cos (3π/4) + i sin (3π/4))
∴ Polar form of (1 +2i) / (1 – 3i) is 1/√2 (cos(3π/4) + i sin (3π/4))
(vii) sin 120o –i cos 120o
Given: Z = sin 120o –i cos 120o
= √3/2 – i (-1/2)
= √3/2 + i (1/2)
So now,
|Z| = √(x2 + y2)
= √((√3/2)2 + (1/2)2)
= √(3/4 + 1/4)
= √(4/4)
= √1
= 1
θ = tan-1 (|y|/ |x|)
= tan-1 ((1/2)/ (√3/2))
= tan-1 (1/√3)
Since x > 0, y >0 complex number lies in 1st quadrant and the value of θ is 00≤θ≤900.
θ = π/6
Z = 1 (cos (π/6) + isin (π/6))
∴ Polar form of √3/2 + i (1/2) is 1 (cos (π/6) + i sin(π/6))
(viii) -16 / (1 + i√3)
Given: Z = -16 / (1 +i√3)
Let us multiply anddivide by (1 – i√3), we get
So now,
|Z| = √(x2 + y2)
= √((-4)2 + (4√3)2)
= √(16 + 48)
= √(64)
= 8
θ = tan-1 (|y|/ |x|)
= tan-1 ((4√3) / 4)
= tan-1 (√3)
Since x < 0, y >0 complex number lies in 2nd quadrant and the value of θ is 900≤θ≤1800.
θ = 2π/3
Z = 8 (cos (2π/3) + isin (2π/3))
∴ Polar form of -16 /(1 + i√3) is 8 (cos (2π/3) + i sin (2π/3))
Question - 2 : - Write (i25)3 in polar form.
Answer - 2 : -
Given: Z = (i25)3
= i75
= i74. i
= (i2)37.i
= (-1)37. i
= (-1). i
= – i
= 0 – i
So now,
|Z| = √(x2 + y2)
= √(02 + (-1)2)
= √(0 + 1)
= √1
θ = tan-1 (|y|/ |x|)
= tan-1 (1/ 0)
= tan-1 ∞
Since x ≥ 0, y < 0complex number lies in 4th quadrant and the value of θ is -900≤θ≤00.
θ = -π/2
Z = 1 (cos (-π/2) + isin (-π/2))
= 1 (cos (π/2) – i sin(π/2))
∴ Polar form of (i25)3 is1 (cos (π/2) – i sin (π/2))
Express the following complex numbers in the form r (cos θ + i sinθ):
(i) 1 + i tan α
(ii) tan α – i
(iii) 1 – sin α + i cos α
(iv) (1 – i) / (cos π/3 + i sin π/3)
Answer - 3 : -
(i) 1 + i tan α
Given: Z = 1 + i tan α
We know that the polarform of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ)
Where,
|Z| = modulus of complexnumber = √(x2 +y2)
θ = arg (z) = argumentof complex number = tan-1 (|y| / |x|)
We also know that tanα is a periodic function with period π.
So α is lying in theinterval [0, π/2) ∪ (π/2, π].
Let us consider case1:
α ∈ [0, π/2)
So now,
|Z| = r = √(x2 + y2)
= √(12 + tan2 α)
= √( sec2 α)
= |sec α| since, sec αis positive in the interval [0, π/2)
θ = tan-1 (|y|/ |x|)
= tan-1 (tanα / 1)
= tan-1 (tanα)
= α since, tan α ispositive in the interval [0, π/2)
∴ Polar form is Z = secα (cos α + i sin α)
Let us consider case2:
α ∈ (π/2, π]
So now,
|Z| = r = √(x2 + y2)
= √(12 + tan2 α)
= √( sec2 α)
= |sec α|
= – sec α since, sec αis negative in the interval (π/2, π]
θ = tan-1 (|y|/ |x|)
= tan-1 (tanα / 1)
= tan-1 (tanα)
= -π + α since, tan αis negative in the interval (π/2, π]
θ = -π + α [since, θlies in 4th quadrant]
Z = -sec α (cos (α –π) + i sin (α – π))
∴ Polar form is Z =-sec α (cos (α – π) + i sin (α – π))
(ii) tan α – i
Given: Z = tan α – i
We know that the polarform of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ)
Where,
|Z| = modulus ofcomplex number = √(x2 +y2)
θ = arg (z) = argumentof complex number = tan-1 (|y| / |x|)
We also know that tanα is a periodic function with period π.
So α is lying in theinterval [0, π/2) ∪ (π/2, π].
Let us consider case1:
α ∈ [0, π/2)
So now,
|Z| = r = √(x2 + y2)
= √(tan2 α + 12)
= √( sec2 α)
= |sec α| since, sec αis positive in the interval [0, π/2)
= sec α
θ = tan-1 (|y|/ |x|)
= tan-1 (1/tanα)
= tan-1 (cotα) since, cot α is positive in the interval [0, π/2)
= α – π/2 [since, θlies in 4th quadrant]
Z = sec α (cos (α –π/2) + i sin (α – π/2))
∴ Polar form is Z = secα (cos (α – π/2) + i sin (α – π/2))
Let us consider case2:
α ∈ (π/2, π]
So now,
|Z| = r = √(x2 + y2)
= √(tan2 α + 12)
= √( sec2 α)
= |sec α|
= – sec α since, sec αis negative in the interval (π/2, π]
θ = tan-1 (|y|/ |x|)
= tan-1 (1/tanα)
= tan-1 (cotα)
= π/2 + α since, cot αis negative in the interval (π/2, π]
θ = π/2 + α [since, θlies in 3th quadrant]
Z = -sec α (cos (π/2 +α) + i sin (π/2 + α))
∴ Polar form is Z =-sec α (cos (π/2 + α) + i sin (π/2 + α))
(iii) 1 – sin α + icos α
Given: Z = 1 – sin α +i cos α
By using the formulas,
Sin2 θ+ cos2 θ = 1
Sin 2θ = 2 sin θ cos θ
Cos 2θ = cos2 θ– sin2 θ
So,
z= (sin2(α/2)+ cos2(α/2) – 2 sin(α/2) cos(α/2)) + i (cos2(α/2)– sin2(α/2))
= (cos(α/2) –sin(α/2))2 + i (cos2(α/2) – sin2(α/2))
We know that the polarform of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ)
Where,
|Z| = modulus ofcomplex number = √(x2 +y2)
θ = arg (z) = argumentof complex number = tan-1 (|y| / |x|)
Now,
We know that sine andcosine functions are periodic with period 2π
Here we have 3intervals:
0 ≤ α ≤ π/2
π/2 ≤ α ≤ 3π/2
3π/2 ≤ α ≤ 2π
Let us consider case1:
In the interval 0 ≤ α≤ π/2
Cos (α/2) > sin(α/2) and also 0 < π/4 + α/2 < π/2
So,
∴ Polar form is Z= √2 (cos (α/2) – sin(α/2)) (cos (π/4 + α/2) + i sin (π/4 + α/2))
Let us consider case2:
In the interval π/2 ≤α ≤ 3π/2
Cos (α/2) < sin(α/2) and also π/2 < π/4 + α/2 < π
So,
Since, (1 – sin α)> 0 and cos α < 0 [Z lies in 4th quadrant]
= α/2 – 3π/4
∴ Polar form is Z = –√2 (cos (α/2) – sin (α/2)) (cos (α/2 –3π/4) + i sin (α/2 – 3π/4))
Let us consider case3:
In the interval 3π/2 ≤α ≤ 2π
Cos (α/2) < sin(α/2) and also π < π/4 + α/2 < 5π/4
So,
θ = tan-1 (tan(π/4 + α/2))
= π – (π/4 + α/2)[since, θ lies in 1st quadrant and tan’s period is π]
= α/2 – 3π/4
∴ Polar form is Z = –√2 (cos (α/2) – sin (α/2)) (cos (α/2 –3π/4) + i sin (α/2 – 3π/4))
(iv) (1 – i) / (cos π/3 + isin π/3)
Given: Z = (1 – i) /(cos π/3 + i sin π/3)
Let us multiply anddivide by (1 – i√3), we get
We know that the polarform of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ)
Where,
|Z| = modulus ofcomplex number = √(x2 +y2)
θ = arg (z) = argumentof complex number = tan-1 (|y| / |x|)
Now,
Since x < 0, y <0 complex number lies in 3rd quadrant and the value of θ is 1800≤θ≤-900.
= tan-1 (2+ √3)
= -7π/12
Z = √2 (cos (-7π/12) + i sin (-7π/12))
= √2 (cos (7π/12) – i sin (7π/12))
∴ Polar form of (1 – i)/ (cos π/3 + i sin π/3) is √2(cos (7π/12) – i sin (7π/12))
Question - 4 : - If z1 and z2 are two complex number suchthat |z1| = |z2| and arg (z1) + arg (z2)= π, then show that 
Answer - 4 : -
|z1| = |z2|and arg (z1) + arg (z2) = π
Let us assume arg (z1)= θ
arg (z2) =π – θ
We know that in thepolar form, z = |z| (cos θ + i sin θ)
z1 =|z1| (cos θ + i sin θ) …………. (i)
z2 =|z2| (cos (π – θ) + i sin (π – θ))
= |z2|(-cos θ + i sin θ)
= – |z2|(cos θ – i sin θ)
Now let us find theconjugate of
= – |z2|(cos θ + i sin θ) …… (ii) (since, |Z2¯¯¯¯¯¯|=|Z2|)
Now,
z1 /= [|z1| (cos θ + i sinθ)] / [-|z2| (cos θ + i sin θ)]
= – |z1| /|z2| [since, |z1| = |z2|]
= -1
When we cross multiplywe get,
z1 = –
Question - 5 : - If z1, z2 and z3, z4 aretwo pairs of conjugate complex numbers, prove that arg (z1/z4)+ arg (z2/z3) = 0
Answer - 5 : -
Given:
Hence proved.
Question - 6 : - Express sin π/5 + i (1 – cos π/5) in polar form.
Answer - 6 : -
Given:
Z = sin π/5 + i (1 –cos π/5)
By using the formula,
sin 2θ = 2 sin θ cos θ
1- cos 2θ = 2 sin2 θ
So,
Z = 2 sin π/10 cosπ/10 + i (2 sin2 π/10)
= 2 sin π/10 (cos π/10+ i sin π/10)
∴ The polar form of sinπ/5 + i (1 – cos π/5) is 2 sin π/10 (cos π/10 + i sin π/10)