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

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

Find the zeros of (z+1)^2+2=0

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

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(z + 1)^{2} + 2 = 0

z = 2 i - 1, z = - 2 i - 1

Solution steps

(z + 1)^{2} + 2 = 0

Move

2

to the right side

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

For

(g (x))^{2} = f (a)

the solutions are

g (x) = f (a), - f (a)

Solve

z + 1 = - 2 : z = 2 i - 1

Solve

z + 1 = - - 2 : z = - 2 i - 1

The solutions to the quadratic equation are:

z = 2 i - 1, z = - 2 i - 1

(x - 2)^{2} = 4 - 9 x

A = 5 x (3 x + 2)

- 7 x^{2} + 4 x + \frac{2}{5} = 0

(x - 2)^{2} - 72 = 0

x^{2} = 5 (x + 210)

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