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

The answer to ((a+bi)+2)^2=-2i is

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((a+bi)+2)^2=-2i

((a + bi) + 2)^{2} = - 2 i

(a = - 3, a = - 1, b = 1 b = - 1)

Solution steps

((a + bi) + 2)^{2} = - 2 i

Expand

((a + bi) + 2)^{2} : (a^{2} + 4 a - b^{2} + 4) + i (4 b + 2 ab)

(a^{2} + 4 a - b^{2} + 4) + i (4 b + 2 ab) = - 2 i

Rewrite

- 2 i

in standard complex form:

0 - 2 i

(a^{2} + 4 a - b^{2} + 4) + i (4 b + 2 ab) = 0 - 2 i

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

[a^{2} + 4 a - b^{2} + 4 = 0 4 b + 2 ab = - 2]

[a^{2} + 4 a - b^{2} + 4 = 0 4 b + 2 ab = - 2] : (a = - 3, a = - 1, b = 1 b = - 1)

(a = - 3, a = - 1, b = 1 b = - 1)

z^{2} + 4 i z - 1 = 0

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

z^{2} = - 15 + 8 i

\frac{( - 6 i + n - 3 n + 5 )}{4 + n ^{2}} = 0

(a + 2 i) (1 + 2 i) = 4 - bi