What are the solutions to the equation 1-z+z^2=0 ?

The solutions to the equation 1-z+z^2=0 are z= 1/2+i(sqrt(3))/2 ,z= 1/2-i(sqrt(3))/2

Find the zeros of 1-z+z^2=0

The zeros of 1-z+z^2=0 are z= 1/2+i(sqrt(3))/2 ,z= 1/2-i(sqrt(3))/2

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1 - z + z^{2} = 0

z = \frac{1}{2} + i \frac{3}{2}, z = \frac{1}{2} - i \frac{3}{2}

Solution steps

1 - z + z^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

z^{2} - z + 1 = 0

Solve with the quadratic formula

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

Simplify

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

z = \frac{1}{2} + i \frac{3}{2}, z = \frac{1}{2} - i \frac{3}{2}

x^{3} (3 x - 5) (4 x + 9) > 0

4 (x + 2)^{2} + 320 = 0

3 (8 x - 6) + 9 = 63

- 9 (\frac{1}{2})

\frac{[ 4 ( 1 + h ) - ( 1 + h ) ^{2} ] - 3}{h}

What are the solutions to the equation 1-z+z^2=0 ?
The solutions to the equation 1-z+z^2=0 are z= 1/2+i(sqrt(3))/2 ,z= 1/2-i(sqrt(3))/2
Find the zeros of 1-z+z^2=0
The zeros of 1-z+z^2=0 are z= 1/2+i(sqrt(3))/2 ,z= 1/2-i(sqrt(3))/2