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Question
Write in polar form of the following complex numbers
– 2 – i2
Solution
– 2 – i2 = r(cos θ + i sin θ)
Let z = – 2 – i2 = r(cos θ + i sin θ)
Equating real and imaginary parts
r cos θ = – 2
r sin θ = – 2
r2 cos2θ + r2 sin2θ = (– 2)2 + (– 2)2
r2 = 4 + 4 = 8
r2 = 8
|z| = r = `2sqrt(2)`
cos θ = `(-2)/(2sqrt(2) = (-1)/sqrt(2)`
sin θ = `(-2)/(2sqrt(2)) = (-1)/sqrt(2)`
Since cos θ and sin θ both are in – ve so lies in III quadrant.
Argument = `2"k"pi - 3 pi/4`
As θ = `pi/4 - pi = - (3pi)/4`
∴ Polar from z = r(cos θ + i sin θ)
– 2 – i2 = `2sqrt(2) (cos(2"k"pi - (3pi)/4) + "i" sin(2"k"pi - 3/4)) "k" ∈ "z"`
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