What is the answer to x^{(2)}+y^{(2)}-2+(x+3y-2)i=0 ?

The answer to x^{(2)}+y^{(2)}-2+(x+3y-2)i=0 is

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x^{(2)}+y^{(2)}-2+(x+3y-2)i=0

x^{(2)} + y^{(2)} - 2 + (x + 3 y - 2) i = 0

(x = - 1, x = \frac{7}{5}, y = 1 y = \frac{1}{5})

Solution steps

x^{(2)} + y^{(2)} - 2 + (x + 3 y - 2) i = 0

Rewrite

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

in standard complex form:

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

(x^{2} + y^{2} - 2) + (x + 3 y - 2) i = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} + y^{2} - 2) + (x + 3 y - 2) i = 0 + 0 i

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

[x^{2} + y^{2} - 2 = 0 x + 3 y - 2 = 0]

[x^{2} + y^{2} - 2 = 0 x + 3 y - 2 = 0] : (x = - 1, x = \frac{7}{5}, y = 1 y = \frac{1}{5})

(x = - 1, x = \frac{7}{5}, y = 1 y = \frac{1}{5})

\frac{1}{z} = \frac{- 8 + 2 i}{2 i}

x^{2} + 5 x + 2 x = 1 - 5 i

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

a, z = (a + bi) (a - bi)

\frac{Z - i}{Z} = 2 + i

What is the answer to x^{(2)}+y^{(2)}-2+(x+3y-2)i=0 ?
The answer to x^{(2)}+y^{(2)}-2+(x+3y-2)i=0 is