# 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

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by sweetysiri92

2015-05-18T11:07:31+05:30

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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

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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
2015-05-18T23:28:10+05:30
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