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

The answer to (2x-3iy)(i+3)^3=2(3+i) is x= 2/25 ,y= 1/25

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

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

x = \frac{2}{25}, y = \frac{1}{25}

Solution steps

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

Expand

(36 x + 78 y) + i (52 x - 54 y) = 6 + 2 i

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

[36 x + 78 y = 6 52 x - 54 y = 2]

[36 x + 78 y = 6 52 x - 54 y = 2] : x = \frac{2}{25}, y = \frac{1}{25}

x = \frac{2}{25}, y = \frac{1}{25}

(1 + i)^{2} - a (1 + i) + b = 0

(4 + i) (x + i y) = (4 - i) (x - i y)

3 z - 2 w = - 5 + 10 i

x^{2} + (6 + 2 i) x + (11 + 10 i) = 0

(3 - i) z + (1 + 2 i) = 7 + i es

What is the answer to (2x-3iy)(i+3)^3=2(3+i) ?
The answer to (2x-3iy)(i+3)^3=2(3+i) is x= 2/25 ,y= 1/25