Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 2 Complex Numbers Ex 2.7

Question 1.
Write in polar form of the following complex numbers.
(i) 2 + i2 √3
Solution:
Let 2 + i2√3 = r (cos θ + i sin θ)
x = 2, y = 2√3 r = \(\sqrt{x^{2}+y^{2}}=\sqrt{4+12}\) = 4
2 = 4 cos θ ⇒ cos θ = \(\frac{1}{2}\)
2√3 = 4 sin θ ⇒ sin θ = \(\frac{√3}{2}\)
⇒ θ is in Ist quadrant
∴ θ = α ⇒ θ = \(\frac{π}{3}\)
Polar form is 4 [cos \(\frac{π}{3}\) + i sin \(\frac{π}{3}\)]
4 [cos (2k + \(\frac{π}{3}\)) + i sin(2k +\(\frac{π}{3}\)) k ∈ z

Samacheer Kalvi 12th Maths Guide

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7

(ii) 3 – i √3
Solution:
Let 3 – i √3 = r (cos θ + i sin θ)
x = 3, y = -√3
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 1
⇒ θ is in the 4th quadrant
∴ θ = -α = (-\(\frac{π}{6}\))
Polar form is 2√3 [cos(-\(\frac{π}{3}\)) + i sin (-\(\frac{π}{3}\))]
2 √3 [cos (2kπ – \(\frac{π}{3}\)) + i sin (2kπ – \(\frac{π}{3}\))] k ∈ z

(iii) -2 – i 2 = r (cos θ + i sin θ)
Solution:
x = 2, y = √2 r = \(\sqrt{x^{2}+y^{2}}\)
= \(\sqrt{4+4}\) = 2√2
-2 = 2√2 cos θ ⇒ cos θ = \(\frac{-1}{√2}\)
-2 = -2√2 sin θ ⇒ sin θ = \(\frac{-1}{√2}\)
θ is in III quadrant
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 2

(iv) Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 3
Solution:
x= -1 y= 1
∴ r = \(\sqrt{x^{2}+y^{2}}\)
= \(\sqrt{1+1}\) = √2
-1 = √2 cos θ ⇒ cos θ = \(\frac{1}{√2}\)
1 = √2 sin θ ⇒ sin θ = \(\frac{1}{√2}\)
⇒ θ is in II quadrant
θ = π – α = π – \(\frac{π}{4}\) = \(\frac{3π}{4}\)
i – 1 = √2 (cos \(\frac{3π}{4}\) + i \(\frac{3π}{4}\))
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 4

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7

Question 2.
Find the rectangular form of the complex numbers
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 5
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 6

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 7
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 8

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7

Question 3.
\(\left(x_{1}+i y_{1}\right)\left(x_{2}+i y_{2}\right)\left(x_{3}+i y_{3}\right) \cdots\left(x_{n}+i y_{n}\right)=a+i b\), show that
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 9
Solution:
Let (x1 + iy1) = r1 (cos θ1 + i sin θ1)
where r1 = \(\sqrt{x_{1}^{2}+y_{1}^{2}}\) and θ1 = tan-1(\(\frac{y_1}{x_1}\))
Similarly (x2 + iy2) = r2 (cos θ2 + i sin θ2)
where r2 = \(\sqrt{x_{2}^{2}+y_{2}^{2}}\) and θ2 = tan-1(\(\frac{y_2}{x_2}\)) and soon. finally
(xn + iyn) = rn (cos θ2 + i sin θn)
where Rn = \(\sqrt{x_{n}^{2}+y_{n}^{2}}\) and θn
= tan-1 (\(\frac{y_n}{x_n}\))
And also let a +ib = R(cos ∅ + i sin ∅)
where R = \(\sqrt{a^{2}+a^{2}}\) and ∅ = tan-1(\(\frac{b}{a}\))
(x1 + iy1) (x2 + iy2) – (xn + iyn) = (a + ib)
r1(cos θ2 + i sin θ1) r2(cos θ2 + i sin θ2)…
rn (cos θn + i sin θn) = R (cos ∅ + i sin ∅)
⇒ {r1 r2……. rn) = R
[(cos (θ1 + θ2 ……..+ θn) + i sin (θ1 + θ2 ……..+ θn)]
= R [cos ∅ + i sin ∅]
Equating the moduli
r1 r2……… rn = R
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 10

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 11
Solution:
Equating the arguments
θ1 + θ2 + …… θn = ∅ + 2kπ k ∈ z
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 12

Question 4.
Given \(\frac{1+z}{1-z}\) = cos 2θ + i sin 2θ, show that To prove that z = i tan θ.
Solution:
z = i tan θ
Now cos2θ + i sin2θ
= (cos²θ – sin²θ) + 2i sinθ cosθ
= (cosθ + i sin²θ)²
This can be written as
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 13

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7

Question 5.
If cos α + cos β + cos γ = sin α + sin β + sin γ = 0. then show that
(i) cos 3α + cos 3β + cos 3γ = 3 cos (α + β + γ) and
(ii) sin 3α + sin 3β + sin 3γ = 3sin (α + β + γ).
Solution:
cos α + cos β + cos γ = 0 ……….(1)
sin a + sin β + sin γ = 0 ………(2)
(1) + i (2) gives
(cos α + i sin α) + (cos β + i sin β) + (cos γ + i sin γ) = 0
Let a = cos α + i sin α, b = cos β + sin β, c = cos γ + i sin γ
∴ a + b + c = 0 => a³ + b³ + c³ = 3abc …….(1)
Now a³ = (cos α + i sin α)³
= cos³ α + 3i cos²α sin α – 3 cos α sin²α – i sin³α
= cos³ α + 3i(1 – sin²α) sin α – 3 cos α (1 – cos²α – i sin³α)
= (cos³α – 3 cos α + 3 cos³α) + i (3 sin α – 3 sin³α – 3 sin³α)
= (4 cos³α – 3 sin α) + i (3 sin α – 4 sin³α)
= cos α + i sin³α
b³ = cos³ β + i sin³β and
c³ = cos³ γ + i sin³γ
Substitute in (1)
(cos³a + i sin³ α) + (cos³β + i sin³β) + (cos³γ + i sin³γ)
= 3(cos α + 1 sin α) (cos β + 1 sin β) (cos γ + 1 sin γ)
= 3[cos (α + β + γ) + i sin (α + β + γ)]
Equating real and imaginary parts,
cos 3α + cos 3β + cos 3γ = 3cos (α + β + γ)
cos 3α + sin 3β + sin 3γ = 3sin (α + β + γ)

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7

Question 6.
If z = x + iy and arg \(\left(\frac{z-i}{z+2}\right)\) = \(\)\frac{π}{4}, then show that x² + y³ + 3x – 3y + 2 = 0.
Solution:
Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7 14
⇒ (y – 1) (x + 2) – xy = x (x + 2) + y (y – 1) xy – x + 2y – 2 – xy = x² + 2x + y² – y
⇒ x² + y² + 3x – 3y + z = 0

Samacheer Kalvi 12th Maths Guide Chapter 2 Complex Numbers Ex 2.7

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