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

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

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

(x = - 2, x = 2, y = 1 y = - 1)

Solution steps

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

Expand

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

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

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

[x^{2} - y^{2} = 3 2 x y = - 4]

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

(x = - 2, x = 2, y = 1 y = - 1)

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

\frac{9}{2} x + (3 i) y = 0

y, (- 2 + 5 i) (1 - i) + (4 + 3 i) = - 6 x + 2 y - 3

x^{2} - 4 x + 4 i = - 1

x + 3 y i = 4 - 5 i