Write the complex number in polar form with argument θ and 2π
a) -2+2i b)√2-√2i c)-√3-i d)-5+5√3i
e)3+3√3i f)2i g)-5i f)-3 h)√2
Answers
Answered by
2
complex numbers - polar form
x + i y = r Cos Ф + i r Sin Ф
=> Tan Ф = y/x and r = √ [ x² + y² ]
a) -2 + 2 i = r (cos Ф + i Sin Ф)
=> Tan Ф = 2/-2 = -1 and r = √[ (-2)² + 2² ] = 2√2
Ф = 3π/4 as cosФ < 0 and sin Ф >0
So 2√2 [ Cos 3π/4 + Sin 3π/4 ]
b) Ф = tan⁻¹ (-√2/√2) = - π/4 as cos Ф >0 and sin Ф < 0
r = √(2+2) = 2
c) r = √ [ 3 + 1 ] = 2
Tan Ф = 1/√3 => Ф = 5π/6 as both cos and sin are -ve
d) - 5 + 5 √3 i
Tan Ф = 5√3/-5 => Ф = 2π/3 as Cos is negative
r = √ [ (-5)² + (5√3)² ] = 10
e) r = √ [ 3² + (3√3)² ] = 6
tan Ф = 3√3/3 => Ф = π/3
f) 2 i => tan Ф = ∞ => Ф = π/2
r = 2
g) - 5 i => tan Ф = ∞ => Ф = - π/2 or 3π/2 as sin is -ve
r = 5
h) √2 => tan Ф = 0 / √2 = 0 => Ф = 0 as cos is +ve
r = √2
x + i y = r Cos Ф + i r Sin Ф
=> Tan Ф = y/x and r = √ [ x² + y² ]
a) -2 + 2 i = r (cos Ф + i Sin Ф)
=> Tan Ф = 2/-2 = -1 and r = √[ (-2)² + 2² ] = 2√2
Ф = 3π/4 as cosФ < 0 and sin Ф >0
So 2√2 [ Cos 3π/4 + Sin 3π/4 ]
b) Ф = tan⁻¹ (-√2/√2) = - π/4 as cos Ф >0 and sin Ф < 0
r = √(2+2) = 2
c) r = √ [ 3 + 1 ] = 2
Tan Ф = 1/√3 => Ф = 5π/6 as both cos and sin are -ve
d) - 5 + 5 √3 i
Tan Ф = 5√3/-5 => Ф = 2π/3 as Cos is negative
r = √ [ (-5)² + (5√3)² ] = 10
e) r = √ [ 3² + (3√3)² ] = 6
tan Ф = 3√3/3 => Ф = π/3
f) 2 i => tan Ф = ∞ => Ф = π/2
r = 2
g) - 5 i => tan Ф = ∞ => Ф = - π/2 or 3π/2 as sin is -ve
r = 5
h) √2 => tan Ф = 0 / √2 = 0 => Ф = 0 as cos is +ve
r = √2
kvnmurty:
please click on thanks blue button above
Sir Below one is copied ur answer
in c) , there is a small mistake. tan π/6 = 1/√3. As Sine and Cosines are both negative, the angle is in 3rd quadrant. So Ф = π + π/6 = 7π/6
Answered by
0
complex numbers - polar form
x + i y = r Cos Ф + i r Sin Ф
=> Tan Ф = y/x and r = √ [ x² + y² ]
a) -2 + 2 i = r (cos Ф + i Sin Ф)
=> Tan Ф = 2/-2 = -1 and r = √[ (-2)² + 2² ] = 2√2
Ф = 3π/4 as cosФ < 0 and sin Ф >0
So 2√2 [ Cos 3π/4 + Sin 3π/4 ]
b) Ф = tan⁻¹ (-√2/√2) = - π/4 as cos Ф >0 and sin Ф < 0
r = √(2+2) = 2
c) r = √ [ 3 + 1 ] = 2
Tan Ф = 1/√3 => Ф = 5π/6 as both cos and sin are -ve
d) - 5 + 5 √3 i
Tan Ф = 5√3/-5 => Ф = 2π/3 as Cos is negative
r = √ [ (-5)² + (5√3)² ] = 10
e) r = √ [ 3² + (3√3)² ] = 6
tan Ф = 3√3/3 => Ф = π/3
f) 2 i => tan Ф = ∞ => Ф = π/2
r = 2
g) - 5 i => tan Ф = ∞ => Ф = - π/2 or 3π/2 as sin is -ve
r = 5
h) √2 => tan Ф = 0 / √2 = 0 => Ф = 0 as cos is +ve
x + i y = r Cos Ф + i r Sin Ф
=> Tan Ф = y/x and r = √ [ x² + y² ]
a) -2 + 2 i = r (cos Ф + i Sin Ф)
=> Tan Ф = 2/-2 = -1 and r = √[ (-2)² + 2² ] = 2√2
Ф = 3π/4 as cosФ < 0 and sin Ф >0
So 2√2 [ Cos 3π/4 + Sin 3π/4 ]
b) Ф = tan⁻¹ (-√2/√2) = - π/4 as cos Ф >0 and sin Ф < 0
r = √(2+2) = 2
c) r = √ [ 3 + 1 ] = 2
Tan Ф = 1/√3 => Ф = 5π/6 as both cos and sin are -ve
d) - 5 + 5 √3 i
Tan Ф = 5√3/-5 => Ф = 2π/3 as Cos is negative
r = √ [ (-5)² + (5√3)² ] = 10
e) r = √ [ 3² + (3√3)² ] = 6
tan Ф = 3√3/3 => Ф = π/3
f) 2 i => tan Ф = ∞ => Ф = π/2
r = 2
g) - 5 i => tan Ф = ∞ => Ф = - π/2 or 3π/2 as sin is -ve
r = 5
h) √2 => tan Ф = 0 / √2 = 0 => Ф = 0 as cos is +ve
Similar questions