What is the answer to (z+2)^2=-2i ?

The answer to (z+2)^2=-2i is z=-1-i,z=-3+i

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(z+2)^2=-2i

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

z = - 1 - i, z = - 3 + i

Solution steps

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

Substitute

z = x + y i

(x + y i + 2)^{2} = - 2 i

Expand

(x + y i + 2)^{2} : (x^{2} + 4 x - y^{2} + 4) + i (4 y + 2 x y)

(x^{2} + 4 x - y^{2} + 4) + i (4 y + 2 x y) = - 2 i

Rewrite

- 2 i

in standard complex form:

0 - 2 i

(x^{2} + 4 x - y^{2} + 4) + i (4 y + 2 x y) = 0 - 2 i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[x^{2} + 4 x - y^{2} + 4 = 0 4 y + 2 x y = - 2]

[x^{2} + 4 x - y^{2} + 4 = 0 4 y + 2 x y = - 2] : (x = - 1, x = - 3, y = - 1 y = 1)

Substitute back

z = x + y i

z = - 1 - i, z = - 3 + i

∣ z - 1 + i ∣ = 1

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

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

2 y + i y = 3 i + i x - x

x^{2} - 4 y + (2 y - x) i = 2 - i