Convert each of the complex numbers given in Exercises 3 to 8 in the polar form:

– 3


Let – 3 = r cos θ and 0 = r sin θ ……….(i)


Squaring both sides, we get


9= r2 cos2 θ and 0 = r2 sin2 θ


Adding both the equations, we get


9 + 0 = r2 cos2 θ + r2 sin2 θ


9 = r2 (cos2 θ + sin2 θ)


9 = r2 or r2 = 9 [ sin2 θ + cos2 θ = 1]


r = 9


r = 3 (conventionally, r>0)


Substituting r = √2 in (i), we get


– 3 = 3 cos θ and 0 = 3 sin θ


cos θ = 1 and sin θ = 0


θ = cos-1 (-1) and θ = sin-1 0


We know that the complex number 3 lies on the real axis (x-axis) and the value of the argument lies between - π and π, i.e. - π < θ ≤ π.


θ = cos-1 cos π and θ = sin-1 sin π


θ = π


As we know that the polar representation of a complex number z = x + iy is


z = r (cos θ + i sin θ) where is the modulus of the complex number and θ is the argument of the complex number, denoted by arg z.


So, the required polar form is z = 3 (cos π + i sin π)


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