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Question
Write in polar form of the following complex numbers
`3 - "i"sqrt(3)`
Solution
Let z = `3 - "i"sqrt(3)` = r(cos θ + i sin θ)
Equating real and imaginary parts
r cos θ = 3(+ve)
r sin θ = `- sqrt(3)` (– ve)
r2 cos2θ + r2 sin2θ = `(3)^2 + (- sqrt(3))^2`
r2 = 9 + 3 = 12
|z| = r = `2sqrt(3)`
Since cos θ positive and sin θ in – ve so lies in IV quadrant.
cos θ = `sqrt(3)/2`
sin θ = `(-1)/2`
θ = `(- pi)/6`
Argument = `2"k"pi - pi/6`, k ∈ z
∴ Polar from z = r(cos θ + i sin θ)
`3 - "i"sqrt(3) = 2sqrt(3) (cos(2"k"pi - pi/6) + "i" sin(2"k"pi - pi/6)) "k" ∈ "z"`
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