What are the solutions to the equation (z+1)^2=-4 ?

The solutions to the equation (z+1)^2=-4 are z=2i-1,z=-2i-1

Find the zeros of (z+1)^2=-4

The zeros of (z+1)^2=-4 are z=2i-1,z=-2i-1

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(z + 1)^{2} = - 4

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

Solution steps

(z + 1)^{2} = - 4

For

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

the solutions are

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

Solve

z + 1 = - 4 : z = 2 i - 1

Solve

z + 1 = - - 4 : z = - 2 i - 1

The solutions to the quadratic equation are:

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

lo g_{\frac{1}{4}} (4)

2 x^{2} + 30 x - 38 = 0

(x - 5)^{2} \leq 0

144 x^{2} - 1

x^{6} + 16 x^{3} + 64

What are the solutions to the equation (z+1)^2=-4 ?
The solutions to the equation (z+1)^2=-4 are z=2i-1,z=-2i-1
Find the zeros of (z+1)^2=-4
The zeros of (z+1)^2=-4 are z=2i-1,z=-2i-1