What is the polar foam of " -3 " ??
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Given: z=a+bi
r=√a^2+b^2
θ=π+tan ^ −1(b / a)
(because it is quadrant 4)
z=r(cos(θ)+isin(θ)
z = -3 + 0.i
r = √(-3)^2 = 3
θ = π + tan ^-1 (0) = π
z = 3( cosπ + isinπ )
r=√a^2+b^2
θ=π+tan ^ −1(b / a)
(because it is quadrant 4)
z=r(cos(θ)+isin(θ)
z = -3 + 0.i
r = √(-3)^2 = 3
θ = π + tan ^-1 (0) = π
z = 3( cosπ + isinπ )
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