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

The answer to (x+iy)^2-3i(x+iy)-2=0 is

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(x+iy)^2-3i(x+iy)-2=0

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

(x = 0, x = 0, y = 1 y = 2)

Solution steps

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

(x = 0, x = 0, y = 1 y = 2)

2 + i y = x - 2 i

\frac{6}{x + y i + ( 2 - i ) ^{2}} = 4 - 3 i

∣ z - i ∣ = 2 z + i

3 i z^{2} - 2 z = - i

x, s i f = y x^{2} - x y^{3}

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