What is the answer to (3x+2yi)^2=(200)/(4+3i) ?

The answer to (3x+2yi)^2=(200)/(4+3i) is

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(3x+2yi)^2=(200)/(4+3i)

(3 x + 2 y i)^{2} = \frac{200}{4 + 3 i}

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

Solution steps

(3 x + 2 y i)^{2} = \frac{200}{4 + 3 i}

Expand

(9 x^{2} - 4 y^{2}) + 12 i x y = 32 - 24 i

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

[9 x^{2} - 4 y^{2} = 32 12 x y = - 24]

[9 x^{2} - 4 y^{2} = 32 12 x y = - 24] : (x = - 2, x = 2, y = 1 y = - 1)

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

2 (5 x + 9 y) - (9 y - 16 x) i = - 11 - 5 i

z^{2} = 2 + 2 i

2 a + bi = - 25 - 2

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

2 z^{2} + (- 1 + i) z + 3 + i = 0

What is the answer to (3x+2yi)^2=(200)/(4+3i) ?
The answer to (3x+2yi)^2=(200)/(4+3i) is