Given complex number -
z=i
R e a l ( z ) = x = 0 I m g ( z ) = y = 1 Real(z)=x=0\\
Img(z)=y=1 R e a l ( z ) = x = 0 I m g ( z ) = y = 1
ϕ = a r g ( z ) = t a n − 1 ( 1 0 ) = t a n − 1 ( ∞ ) = π 2 \phi=arg(z)=tan^{-1}(\dfrac{1}{0})=tan^{-1}(\infty)=\dfrac{\pi}{2} ϕ = a r g ( z ) = t a n − 1 ( 0 1 ) = t a n − 1 ( ∞ ) = 2 π
The above complex number in polar form is-
z= c o s π 2 + i s i n π 2 =cos\dfrac{\pi}{2}+isin\dfrac{\pi}{2} = cos 2 π + i s in 2 π
To find its cube root we take-
z k = z 3 = ∣ z ∣ 3 ( c o s ϕ + 2 π k 3 + i s i n ϕ + 2 π k 3 ) , k = 0 , 1 , 2. z_k=\sqrt[3]{z}=\sqrt[3]{|z|}(cos {\frac {\phi+2\pi k} 3}+isin{\frac {\phi+2\pi k} 3}), k=0,1,2. z k = 3 z = 3 ∣ z ∣ ( cos 3 ϕ + 2 πk + i s in 3 ϕ + 2 πk ) , k = 0 , 1 , 2.
So, let's find all z k z_k z k
z 0 = ( c o s π 6 + i s i n π 6 ) z_0=(cos {\frac {\pi} 6}+isin{\frac {\pi} 6}) z 0 = ( cos 6 π + i s in 6 π )
z 1 = ( c o s 5 π 6 + i s i n 5 π 6 ) z_1=(cos {\frac {5\pi} 6}+isin{\frac {5\pi} 6}) z 1 = ( cos 6 5 π + i s in 6 5 π )
z 2 = ( c o s 3 π 2 + i s i n 3 π 2 ) z_2=(cos {\frac {3\pi} 2}+isin{\frac {3\pi} 2}) z 2 = ( cos 2 3 π + i s in 2 3 π )
Here Demovier's theorem is used.
Argand diagram is-
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