θ sin θ cos θ |
0 0 1 |
30 1/2 √3/2 |
45 1/√2 1/√2 |
60 √3/2 1/2 |
90 1 0 |
Property 1 :
If z = r (cos θ + i sin θ), then
z-1 = (1/r) (cos θ - i sin θ)
Property 2 :
If z1 = r1 (cos θ1 + i sin θ1) and z2 = r2 (cos θ2 + i sin θ2) then
z1 z2 = r1r2 (cos (θ1 + θ2) + i sin (θ1 + θ2))
Property 3 :
If z1 = r1 (cos θ1 + i sin θ1) and z2 = r2 (cos θ2 + i sin θ2) then
z1/ z2 = (r1/r2) (cos (θ1 - θ2) + i sin (θ1 - θ2))
Find the rectangular form of the following complex numbers :
Problem 1 :
[cos (π/6) + i sin (π/6)] [cos (π/12) + i sin (π/12)]
Solution :
= [cos (π/6) + i sin (π/6)] [cos (π/12) + i sin (π/12)]
= [cos ((π/6) + (π/12)) + i sin ((π/6) + (π/12))]
= cos (2π + π)/12 + i sin (2π + π)/12
= cos (3π/12) + i sin (3π/12)
= cos (π/4) + i sin (π/4)
= (1/√2) + i (1/√2)
= (1/√2)(1 + i)
Problem 2 :
[cos (π/6) - i sin (π/6)]/2 [cos (π/3) + i sin (π/3)]
Solution :
= [cos (π/6) - i sin (π/6)cos (π/3) + i sin (π/3)]
= (1/2) [cos (-π/6)+sin (-π/6)] [cos (-π/3)+i sin (-π/3)]
= (1/2) [cos (-3π/6) + sin (-3π/6)]
= (1/2) [cos (-π/2) + sin (-π/2)]
= (1/2) [0 + (-1)]
= (-1/2)
Problem 3 :
If (x1 + iy1) (x2 + iy2) (x3 + iy3) ................(xn + iyn) = a + ib show that
(x12 + y12) (x22 + y22)............ (xn2 + yn2) = a2 + b2
Solution :
(x1 + iy1) (x2 + iy2) (x3 + iy3) ................(xn + iyn) = a + ib
Taking modulus on both sides, we get
|(x1 + iy1) (x2 + iy2) (x3 + iy3) ................(xn + iyn)| = |a + ib|
|(x1 + iy1)| |(x2 + iy2)| |(x3 + iy3)|............|(xn + iyn)| = |a + ib|
√(x12 + y12) √(x22 + y22)√(x32 + y32) ...............√(xn2 + y2) = √(a2 + b2)
√[(x12 + y12) (x22 + y22)(x32 + y32) ..............(xn2 + y2)] = √(a2 + b2)
Taking squares on both sides, we get
(x12+y12) (x22 + y22)(x32 + y32) ..............(xn2 + y2) = (a2+b2)
Problem 4 :
If (x1 + iy1) (x2 + iy2) (x3 + iy3) ................(xn + iyn) = a + ib show that
∑r = 1 to n tan-1(yr/xr) = tan-1(b/a) + 2kπ, k ∈ z
Solution :
(x1 + iy1) (x2 + iy2) (x3 + iy3) ................(xn + iyn) = a + ib
By taking argument on both sides, we get
arg [(x1 + iy1) (x2 + iy2) (x3 + iy3) ................(xn + iyn)] = arg(a + ib)
arg(x1+ iy1) + arg(x2+ iy2)+................+ arg(x2+ iy2) = arg(a + ib)
tan-1(y1/x1) + tan-1(y2/x2) + ..................+ tan-1(yn/xn) = tan-1(b/a)
∑r = 1 to n tan-1(yr/xr) = tan-1(b/a) + 2kπ, k ∈ z
Problem 5 :
Find the product
(3/2) (cos π/3 + i sin π/3) ⋅ 6 (cos 5π/6 + i sin 5π/6)
Solution :
= (3/2) (cos π/3 + i sin π/3) ⋅ 6 (cos 5π/6 + i sin 5π/6)
= (3/2) ⋅ 6 (cos π/3 + i sin π/3) (cos 5π/6 + i sin 5π/6)
Looks like,
(cos A + i sin A)(cos B + i sin B) = cos (A + B) + i sin (A + B)
= 3 ⋅ 3 (cos (π/3 + 5π/6) + i sin (π/3 + 5π/6)
= 9 (cos (2π + 5π)/6 + i sin (2π + 5π)/6
= 9 (cos 7π/6 + i sin 7π/6)
= 9 [cos (π + π/6) + i sin (π + π/6)]
Angle lies in third quadrant, then using ASTC, we get
= 9 [-cos (π/6) - i sin (π/6)]
= 9 [-√3/2 - i (1/2)]
= [-9√3/2 - i (9/2)]
Problem 6 :
Find the product
2 (cos 9π/4 + i sin 9π/4) / 4 (cos (-3π/2) + i sin (-3π/2))
Solution :
= 2 (cos 9π/4 + i sin 9π/4) / 4 (cos(-3π/2) + i sin(-3π/2))
= (1/2) ( cos (9π/4 + 3π/2) + i sin (9π/4 + 3π/2) )
= (1/2) ( cos (9π + 6π)/4) + i sin (9π + 6π)/4) )
= (1/2) ( cos (15π/4) + i sin (15π/4) )
= (1/2) ( cos (4π - π/4) + i sin (4π - π/4) )
= (1/2) ( cos (π/4) + i sin (π/4) )
= (1/2) ( √2/2 + i√2/2 )
= (1/4) (√2 + i√2)
Problem 7 :
Find the product
(cos π/6 + i sin π/6) (cos π/12 + i sin π/12)
Solution :
= (cos π/6 + i sin π/6) (cos π/12 + i sin π/12)
= cos (π/6 + π/12) + i sin (π/6 + π/12)
= cos (2π+π)/12 + i sin (2π+π)/12
= cos (3π/12) + i sin (3π/12)
= cos (π/4) + i sin (π/4)
= ( √2/2 + i√2/2 )
Find the corresponding polar coordinates for the given rectangular coordinate where 0 ≤ θ ≤ 2π
Problem 8 :
(-5, -5)
Solution :
Every point will be in the form (x, y)
x = -5 and y = -5
Polar form will be in the form,
= r (cos θ + i sin θ)
r = √(-5)2 + 52
= √25 + 25
= √50
= 5√2
θ = tan-1(y/x)
= tan-1(-5/(-5))
= tan-1(1)
= π/4
So, the required answer is
5√2 (cos π/4 + i sin π/4)
Problem 9 :
(-√3, 1)
Solution :
Every point will be in the form (x, y)
x = -√3 and y = 1
Polar form will be in the form,
= r (cos θ + i sin θ)
r = √(-√3)2 + 12
= √3+1
= √4
= 2
θ = tan-1(y/x)
= tan-1(-1/√3)
= -tan-1(1/√3)
= -π/6
= π - π/6
= 5π/6
So, the required answer is
2 (cos 5π/6 + i sin 5π/6)
Kindly mail your feedback to v4formath@gmail.com
We always appreciate your feedback.
©All rights reserved. onlinemath4all.com
Dec 10, 24 05:46 AM
Dec 10, 24 05:44 AM
Dec 08, 24 08:11 PM