Given, z=−3
Let, rcosθ=−3 and rsinθ=0
On squaring and adding, we obtain
r2cos2θ+r2sin2θ=(−3)2
⇒r2(cos2θ+sin2θ)=9
⇒r2=9
⇒r=√9=3 (Conventionally r>0 )
∴3cosθ=−3 and 3sinθ=0
⇒cosθ=−1 and sinθ=0
∴θ=π
So, the polar form is
∴−3=rcosθ+irsinθ=3cosπ+3sinπ=3(cosπ+isinπ)