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In this article, we will delve into the solutions for Exercise 13.4 from Chapter 13 of RD Sharma's Class 11 Mathematics textbook which covers "Complex Numbers". This chapter introduces students to the fundamental concepts of the complex numbers a crucial area in the algebra that extends the real number system. The solutions provided here will offer step-by-step guidance to help students tackle various problems involving complex numbers enhancing their understanding and problem-solving skills.
Complex numbers are numbers that include a real part and an imaginary part expressed in the form a+bi where a and b are real numbers and i is the imaginary unit with the property i2=−1. They are used to solve equations that cannot be solved using only real numbers. Understanding complex numbers is essential for higher-level mathematics and various applications in engineering and physics.
(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)
The polar form of a complex number Z = x + iy is given by Z = |Z| (cos θ + i sin θ) where,
Modulus of complex number, |Z| = √(x2 + y2)
Argument of complex number, θ = arg (Z) = tan–1 (y/x)
Solution:
We are given, Z = 1 + i, so x = 1 and y = 1.
|Z| = √(12 + 12) = √2
θ = tan-1 (1/1) = tan-1 1
As x > 0 and y > 0, Z lies in 1st quadrant and the value of θ is 0 ≤ θ ≤ π/2.
So, θ = π/4 and Z = √2 (cos (π/4) + i sin (π/4))
Therefore, the polar form of (1 + i) is √2 (cos (π/4) + i sin (π/4)).
Solution:
We are given, Z = √3 + i, so x = √3 and y = 1.
|Z| = √((√3)2 + 12) = 2
θ = tan-1 (1/√3)
As x > 0 and y > 0, Z lies in 1st quadrant and the value of θ is 0 ≤ θ ≤ π/2.
So, θ = π/6 and Z = 2 (cos (π/6) + i sin (π/6))
Therefore, the polar form of (√3 + i) is √2 (cos (π/6) + i sin (π/6)).
Solution:
We are given, Z = 1 – i, so x = 1 and y = –1.
|Z| = √(1)2 + (–1)2) = √2
θ = tan-1 (1/1) = tan-1 1
As x > 0 and y < 0, Z lies in 4th quadrant and the value of θ is –π/2 ≤ θ ≤ 0.
So, θ = – π/4 and,
Z = √2 (cos (–π/4) + i sin (–π/4))
= √2 (cos (π/4) – i sin (π/4))
Therefore, the polar form of (1 – i) is √2 (cos (π/4) – i sin (π/4)).
Solution:
We are given, Z = (1 – i)/(1 + i).
Multiplying and dividing by (1 – i), we get,
Z =
=
=
=
= 0 – i
So x = 0, y = –1 and |Z| = √(02 + (–1)2) = 1
θ = tan-1 (1/0)
As x ≥ 0 and y < 0, Z lies in 4th quadrant and the value of θ is –π/2 ≤ θ ≤ 0.
So, θ = –π/2 and,
Z = 1 (cos (–π/2) + i sin (–π/2))
= cos (π/2) – i sin (π/2)
Therefore, the polar form of (1 – i)/(1 + i) is cos (π/2) – i sin (π/2).
Solution:
We are given, Z = (1 – i)/(1 + i).
Multiplying and dividing by (1 – i), we get,
Z =
=
=
=
= 1/2 – i/2
So x = 1/2, y = –1/2 and |Z| = √((1/2)2 + (–1/2)2) = √(2/4) = 1/√2
θ = tan-1 ((1/2)/(1/2)) = tan–1 1
As x > 0 and y < 0, Z lies in 4th quadrant and the value of θ is –π/2 ≤ θ ≤ 0.
So, θ = –π/4 and,
Z = 1/√2 (cos (-π/4) + i sin (-π/4))
= 1/√2 (cos (π/4) – i sin (π/4))
Therefore, the polar form of 1/(1 + i) is 1/√2 (cos (π/4) – i sin (π/4)).
Solution:
We are given, Z = (1 + 2i)/(1 – 3i).
Multiplying and dividing by (1 + 3i), we get,
Z =
=
=
=
= –1/2 + i/2
So x = –1/2, y = 1/2 and |Z| = √((–1/2)2 + (1/2)2) = √(2/4) = 1/√2
θ = tan-1 ((1/2)/(1/2)) = tan–1 1
As x < 0 and y > 0, Z lies in 2nd quadrant and the value of θ is π/2 ≤ θ ≤ π.
So, θ = 3π/4 and Z = 1/√2 (cos (3π/4) + i sin (3π/4))
Therefore, the polar form of (1 + 2i)/(1 – 3i) is 1/√2 (cos (3π/4) + i sin (3π/4)).
Solution:
We are given, Z = sin 120o – i cos 120o
= √3/2 – i (–1/2)
= √3/2 + i (1/2)
So x = √3/2, y = 1/2 and |Z| = √((√3/2)2 + (1/2)2) = √(3/4 + 1/4) = 1
θ = tan-1 ((1/2)/(√3/2)) = tan-1 (1/√3)
As x > 0 and y > 0, Z lies in 1st quadrant and the value of θ is 0 ≤ θ ≤ π/2.
So, θ = π/6 and Z = 1 (cos (π/6) + i sin (π/6))
Therefore, the polar form of √3/2 + i (1/2) is 1 (cos (π/6) + i sin (π/6)).
We are given, Z = –16/(1 + i√3).
Multiplying and dividing by (1 – i√3), we get,
Z =
=
=
= –4 + 4√3 i
So x = –4, y = 4/√3 and |Z| = √((–4)2 + (4√3)2) = √(16 + 48) = 8
θ = tan-1 (4√3/4) = tan-1 (√3)
As x < 0 and y > 0, Z lies in 2nd quadrant and the value of θ is π/2 ≤ θ ≤ π.
So, θ = 2π/3 and Z = 8 (cos (2π/3) + i sin (2π/3))
Therefore, the polar form of -16 / (1 + i√3) is 8 (cos (2π/3) + i sin (2π/3)).
Solution:
We are given,
Z = (i25)3
= i75
= (i2)37. i
= (–1)37. i
= – i
= 0 – i
So x = 0, y = –1 and,
|Z| = √(x2 + y2)
= √(02 + (–1)2)
= 1
θ = tan–1 (|y| / |x|)
= tan–1 (1 / 0)
Since x ≥ 0 and y < 0, Z lies in 4th quadrant and the value of θ is π/2 ≤ θ ≤ 0. So, θ = –π/2.
Z = 1 (cos (–π/2) + i sin (–π/2))
= 1 (cos (π/2) – i sin (π/2))
Therefore, the polar form of (i25)3 is 1 (cos (π/2) – i sin (π/2)).
(i) 1 + i tan α
(ii) tan α – i
(iii) 1 − sin α + i cos α
(iv)
Solution:
(i) 1 + i tan α
We are given 1 + i tan α, so x =1 and y = tan α.
We also know that tan α is a periodic function with period π.
So α is lying in the interval [0, π/2) ∪ (π/2, π].
Case 1: If α ∈ [0, π/2)
|Z| = r = √(12 + tan2 α)
= √( sec2 α)
= sec α
θ = tan-1 (tan α/1)
= tan-1 (tan α)
= α
So, Z = sec α (cos α + i sin α)
Therefore, polar form is sec α (cos α + i sin α).
Case 2: α ∈ (π/2, π]
|Z| = r = √(12 + tan2 α)
= √( sec2 α)
= – sec α
θ = tan-1 (tan α/1)
= tan-1 (tan α)
= –π + α
So, Z = –sec α (cos (α – π) + i sin (α – π))
Therefore, polar form is –sec α (cos (α – π) + i sin (α – π)).
(ii) tan α – i
We are given tan α – i, so x =tan α and y = –1.
We also know that tan α is a periodic function with period π.
So α is lying in the interval [0, π/2) ∪ (π/2, π].
Case 1: If α ∈ [0, π/2)
|Z| = r = √(tan2 α + 1)
= √( sec2 α)
= sec α
θ = tan-1 (1/tan α)
= tan-1 (cot α)
= α – π/2
So, Z = sec α (cos (α – π/2) + i sin (α – π/2))
Therefore, polar form is sec α (cos (α – π/2) + i sin (α – π/2)).
Case 2: α ∈ (π/2, π]
|Z| = r = √(tan2 α + 1)
= √( sec2 α)
= – sec α
θ = tan-1 (1/tan α)
= tan-1 (cot α)
= π/2 + α
So, Z = –sec α (cos (π/2 + α) + i sin (π/2 + α))
Therefore, polar form is –sec α (cos (π/2 + α) + i sin (π/2 + α)).
(iii) 1 − sin α + i cos α
Let z = 1 − sin α + i cos α
As sine and cosine functions are periodic functions with period 2π, let us take α in [0, 2π].
Now, z = 1 − sin α + i cos α
Let θ be an acute angle given by,
tan θ =
tan θ =
tan θ =
Case 1: When 0 ≤ α < π/2
In this case, we have,
|z| = √2(cos α/2 - sin α/2)
Also,
tan θ = |tan (π/4 + α/2)| = tan (π/4 + α/2)
θ = π/4 + α/2
Clearly, z lies in the first quadrant . Therefore, arg(z) = π/4 + α/2
Therefore, the polar form of z is √2(cos α/2 - sin α/2){cos (π/4 + α/2) + i sin (π/4 + α/2)}.
Case 2: When π/2 < α < 3π/2
In this case, we have,
|z| = |√2(cos α/2 - sin α/2)| = -√2(cos α/2 - sin α/2)
And, tan θ = |tan (π/4 + α/2)| = -tan (π/4 + α/2) = tan {π - (π/4 + α/2)} = tan (α/2 - 3π/4)
θ = 3π/4 + α/2
Clearly, z lies in the fourth quadrant . Therefore, arg(z) = -θ = 3π/4 + α/2 = α/2 - 3π/4
Therefore, the polar form of z is -√2(cos α/2 - sin α/2){cos (α/2 - sin 3π/4) + i sin (α/2 - sin 3π/4)}..
Case 3: When 3π/2 < α < 2π
In this case, we have,
|z| = |√2(cos α/2 - sin α/2)| = -√2(cos α/2 - sin α/2)
And tan θ = |tan (π/4 + α/2)| = tan (π/4 + α/2) = - tan {π - (π/4 + α/2)} = tan (α/2 - 3π/4)
θ = α/2 - 3π/4
Clearly, z lies in the first quadrant . Therefore, arg(z) = θ = α/2 - 3π/4
Therefore, the polar form of z is -√2(cos α/2 - sin α/2){cos (α/2 - sin 3π/4) + i sin (α/2 - sin 3π/4)}.
(iv)
Let z =
=
=
=
=
=
=
Now, z =
=
=
= √2
Let θ be an acute angle given by tan θ =
tan θ =
tan θ = |tan (π/4 + π/3)| = |tan 7π/12|
θ = 7π/12
Clearly, z lies in the fourth quadrant . Therefore, arg(z) = -7π/12
Therefore, the polar form of z is √2(cos 7π/12 - sin 7π/12).
Solution:
We are given |z1| = |z2| and arg (z1) + arg (z2) = π. Suppose arg (z1) = θ, then arg (z2) = π – θ.
We know z = |z| (cos θ + i sin θ)
z2 = |z2| (cos (π – θ) + i sin (π – θ))
= |z2| (–cos θ + i sin θ)
= – |z2| (cos θ – i sin θ)
The conjugate of z2, = – |z2| (cos θ + i sin θ)
Now L.H.S. = z1 = |z1| (cos θ + i sin θ)
= |z2| (cos θ + i sin θ)
= – [– |z2| (cos θ + i sin θ)]
=
= R.H.S.
Hence proved.
Solution:
We are given,
L.H.S. = arg (z1/z4) + arg (z2/z3)
= arg (z1) − arg (z4) + arg (z2) − arg (z3)
= [arg (z1) + arg (z2)] − [arg (z3) + arg (z4)]
=
= 0 − 0
= 0
= R.H.S
Hence proved.
Solution:
We are given,
Z = sin π/5 + i (1 – cos π/5)
= 2 sin π/10 cos π/10 + i (2 sin2 π/10)
= 2 sin π/10 (cos π/10 + i sin π/10)
We know the polar form is given by r (cos θ + i sin θ).
Therefore, the polar form of the given expression is 2 sin π/10 (cos π/10 + i sin π/10).\
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