What is the answer to 1-i=(2-2y+3x)+i(-3+2x+3y) ?

The answer to 1-i=(2-2y+3x)+i(-3+2x+3y) is x= 1/13 ,y= 8/13

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1-i=(2-2y+3x)+i(-3+2x+3y)

1 - i = (2 - 2 y + 3 x) + i (- 3 + 2 x + 3 y)

x = \frac{1}{13}, y = \frac{8}{13}

Solution steps

1 - i = (2 - 2 y + 3 x) + i (- 3 + 2 x + 3 y)

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

[1 = 2 - 2 y + 3 x - 1 = - 3 + 2 x + 3 y]

[1 = 2 - 2 y + 3 x - 1 = - 3 + 2 x + 3 y] : x = \frac{1}{13}, y = \frac{8}{13}

x = \frac{1}{13}, y = \frac{8}{13}

2 x + 4 i y - i x + 5 y = 1 - 7 i

i (x) = x^{2} + x + 1

i (x) = - 100 x^{2} + 1000 x + 7500

- w^{2} + 9 i w + 20 = 0

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

What is the answer to 1-i=(2-2y+3x)+i(-3+2x+3y) ?
The answer to 1-i=(2-2y+3x)+i(-3+2x+3y) is x= 1/13 ,y= 8/13