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

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

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solvefor x,(x+y)^2=3-4i

solve for

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

x = (- y - 2) + i, x = (2 - y) - i

Solution steps

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

Substitute

x = a + bi

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

Expand

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

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

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

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

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

Substitute back

x = a + bi

x = (- y - 2) + i, x = (2 - y) - i

z^{2} = 3 + 3 i

z^{2} = i (1 - 0. 5^{2})

x (1 + i) + y (1 - i) = 0

7 i - 3 j = a (i - j) + b (i + j)

x^{2} + 6 x i - 9 = (5 + 12 i) y

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