What is the answer to (x+4i)^2=3-4i ?

The answer to (x+4i)^2=3-4i is x=2-5i,x=-2-3i

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(x+4i)^2=3-4i

(x + 4 i)^{2} = 3 - 4 i

x = 2 - 5 i, x = - 2 - 3 i

Solution steps

(x + 4 i)^{2} = 3 - 4 i

Substitute

x = a + bi

(a + bi + 4 i)^{2} = 3 - 4 i

Expand

(a + bi + 4 i)^{2} : (a^{2} - 16 - b^{2} - 8 b) + i (8 a + 2 ab)

(a^{2} - 16 - b^{2} - 8 b) + i (8 a + 2 ab) = 3 - 4 i

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

[a^{2} - 16 - b^{2} - 8 b = 3 8 a + 2 ab = - 4]

[a^{2} - 16 - b^{2} - 8 b = 3 8 a + 2 ab = - 4] : (a = 2, a = - 2, b = - 5 b = - 3)

Substitute back

x = a + bi

x = 2 - 5 i, x = - 2 - 3 i

x^{2} + 2 i x y - y^{2} = 0 + 2 i

x - i y = \frac{7 - i ^{9}}{1 - i}

i (x + i y) = 2

i (x) = \frac{x ^{2}}{4} + 3450 x

4 x + 4 i y - 23 = 5 i x + 5 i^{2} y + 2 i

What is the answer to (x+4i)^2=3-4i ?
The answer to (x+4i)^2=3-4i is x=2-5i,x=-2-3i