What are the solutions to the equation z^2+2z=sqrt(i)-1 ?

The solutions to the equation z^2+2z=sqrt(i)-1 are z=-1+\sqrt[4]{i},z=-\sqrt[4]{i}-1

Find the zeros of z^2+2z=sqrt(i)-1

The zeros of z^2+2z=sqrt(i)-1 are z=-1+\sqrt[4]{i},z=-\sqrt[4]{i}-1

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z^{2} + 2 z = i - 1

Solution steps

z^{2} + 2 z = i - 1

Subtract

i - 1

from both sides

z^{2} + 2 z - (i - 1) = i - 1 - (i - 1)

Simplify

z^{2} + 2 z - i + 1 = 0

Solve with the quadratic formula

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

Simplify

Separate the solutions

The solutions to the quadratic equation are:

3 x^{2} - 11 x + 16 = 0

2 w^{2} - 4 w + 4 = 3 w

7 x^{2} - 44 x - 35 = 0

8 x^{2} + 120 = 0

(x - 7) (x + 2) = 3 x^{2} - 5 x - 22

What are the solutions to the equation z^2+2z=sqrt(i)-1 ?
The solutions to the equation z^2+2z=sqrt(i)-1 are z=-1+\sqrt[4]{i},z=-\sqrt[4]{i}-1
Find the zeros of z^2+2z=sqrt(i)-1
The zeros of z^2+2z=sqrt(i)-1 are z=-1+\sqrt[4]{i},z=-\sqrt[4]{i}-1