What are the solutions to the equation z^2=(1-i)^{1/3} ?

The solutions to the equation z^2=(1-i)^{1/3} are z=((1-i)^{1/3})^{1/2},z=-((1-i)^{1/3})^{1/2}

Find the zeros of z^2=(1-i)^{1/3}

The zeros of z^2=(1-i)^{1/3} are z=((1-i)^{1/3})^{1/2},z=-((1-i)^{1/3})^{1/2}

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z^{2} = (1 - i)^{\frac{1}{3}}

z = ((1 - i)^{\frac{1}{3}})^{\frac{1}{2}}, z = - ((1 - i)^{\frac{1}{3}})^{\frac{1}{2}}

Solution steps

z^{2} = (1 - i)^{\frac{1}{3}}

Move

(1 - i)^{\frac{1}{3}}

to the left side

z^{2} - (1 - i)^{\frac{1}{3}} = 0

Solve with the quadratic formula

z_{1, 2} = \frac{- 0 \pm ( 0 ^{2} - 4 \cdot 1 \cdot ( - ( 1 - i ) ^{\frac{1}{3}} ) ) ^{\frac{1}{2}}}{2 \cdot 1}

Simplify

(0^{2} - 4 \cdot 1 \cdot (- (1 - i)^{\frac{1}{3}}))^{\frac{1}{2}} : 2 ((1 - i)^{\frac{1}{3}})^{\frac{1}{2}}

z_{1, 2} = \frac{- 0 \pm 2 ( ( 1 - i ) ^{\frac{1}{3}} ) ^{\frac{1}{2}}}{2 \cdot 1}

Separate the solutions

z_{1} = \frac{- 0 + 2 ( ( 1 - i ) ^{\frac{1}{3}} ) ^{\frac{1}{2}}}{2 \cdot 1}, z_{2} = \frac{- 0 - 2 ( ( 1 - i ) ^{\frac{1}{3}} ) ^{\frac{1}{2}}}{2 \cdot 1}

z = \frac{- 0 + 2 ( ( 1 - i ) ^{\frac{1}{3}} ) ^{\frac{1}{2}}}{2 \cdot 1} : ((1 - i)^{\frac{1}{3}})^{\frac{1}{2}}

z = \frac{- 0 - 2 ( ( 1 - i ) ^{\frac{1}{3}} ) ^{\frac{1}{2}}}{2 \cdot 1} : - ((1 - i)^{\frac{1}{3}})^{\frac{1}{2}}

The solutions to the quadratic equation are:

z = ((1 - i)^{\frac{1}{3}})^{\frac{1}{2}}, z = - ((1 - i)^{\frac{1}{3}})^{\frac{1}{2}}

2 x (x + 7) = 0

5 - x^{2} = 4 x

a^{2} + 2 a - 3 = 0

5 x^{2} + 4 x = 2 x^{2} - 5 x

28 x^{2} - 85 x + 63 = 0

What are the solutions to the equation z^2=(1-i)^{1/3} ?
The solutions to the equation z^2=(1-i)^{1/3} are z=((1-i)^{1/3})^{1/2},z=-((1-i)^{1/3})^{1/2}
Find the zeros of z^2=(1-i)^{1/3}
The zeros of z^2=(1-i)^{1/3} are z=((1-i)^{1/3})^{1/2},z=-((1-i)^{1/3})^{1/2}