Write –3i in polar form.

Let, z = -3i

Let 0 = rcosθ and -3 = rsinθ


By squaring and adding, we get


(0)2 + (-3)2 = (rcosθ)2 + (rsinθ)2


0+9 = r2(cos2θ + sin2θ)


9 = r2


r = 3


cosθ= 0 and sinθ=-1


Since, θ lies in fourth quadrant, we have



Thus, the required polar form is


1