Multiply the numerator and denominator of this complex number by j
z = 1 2 j = 1 2 j ⋅ j j = j 2 j 2 = j 2 ⋅ ( − 1 ) = − 1 2 j z=\frac{1}{2j}=\frac{1}{2j}\cdot \frac{j}{j}=\frac{j}{2{{j}^{2}}}=\frac{j}{2\cdot \left( -1 \right)}=-\frac{1}{2}j z = 2 j 1 = 2 j 1 ⋅ j j = 2 j 2 j = 2 ⋅ ( − 1 ) j = − 2 1 j So in a rectangular form we have
z = x + j y = − 1 2 j z=x+jy=-\frac{1}{2}j z = x + j y = − 2 1 j where
x = 0 , y = − 1 2 x=0,\,\,y=-\frac{1}{2} x = 0 , y = − 2 1 In polar form the complex number z is written as
z = r ( cos θ + j sin θ ) z=r\left( \cos \theta +j\sin \theta \right) z = r ( cos θ + j sin θ ) where
r = ∣ z ∣ = x 2 + y 2 = 0 2 + ( − 1 2 ) 2 = 1 4 = 1 2 r=\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}=\sqrt{{{0}^{2}}+{{\left( -\frac{1}{2} \right)}^{2}}}=\sqrt{\frac{1}{4}}=\frac{1}{2} r = ∣ z ∣ = x 2 + y 2 = 0 2 + ( − 2 1 ) 2 = 4 1 = 2 1 θ = arctan ( y x ) = arctan ( − 1 / 2 0 ) = arctan ( − ∞ ) = − π 2 \theta =\arctan \left( \frac{y}{x} \right)=\arctan \left( \frac{-1/2}{0} \right)=\arctan \left( -\infty \right)=-\frac{\pi }{2} θ = arctan ( x y ) = arctan ( 0 − 1/2 ) = arctan ( − ∞ ) = − 2 π
or otherwise
θ = − π 2 + 2 π = 3 π 2 \theta =-\frac{\pi }{2}+2\pi =\frac{3\pi }{2} θ = − 2 π + 2 π = 2 3 π Substituting the obtained values, we get the polar form of this z
z = 1 2 ( cos ( 3 π 2 ) + j sin ( 3 π 2 ) ) z=\frac{1}{2}\left( \cos \left( \frac{3\pi }{2} \right)+j\sin \left( \frac{3\pi }{2} \right) \right) z = 2 1 ( cos ( 2 3 π ) + j sin ( 2 3 π ) )
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