# 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

2
e)3+3√3i f)2i g)-5i f)-3 h)√2

by sweetysiri92 18.05.2015

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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

2
e)3+3√3i f)2i g)-5i f)-3 h)√2

by sweetysiri92 18.05.2015

Log in to add a comment

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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

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