What is the answer to -(2+3i)z-5+i=-z^2 ?

The answer to -(2+3i)z-5+i=-z^2 is z=-1+i,z=3+2i

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-(2+3i)z-5+i=-z^2

- (2 + 3 i) z - 5 + i = - z^{2}

z = - 1 + i, z = 3 + 2 i

Solution steps

- (2 + 3 i) z - 5 + i = - z^{2}

Substitute

z = x + y i

- (2 + 3 i) (x + y i) - 5 + i = - (x + y i)^{2}

Expand

(- 2 x + 3 y - 5) + i (1 - 2 y - 3 x) = (- x^{2} + y^{2}) - 2 i x y

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

[- 2 x + 3 y - 5 = - x^{2} + y^{2} 1 - 2 y - 3 x = - 2 x y]

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

Substitute back

z = x + y i

z = - 1 + i, z = 3 + 2 i

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

z, 2 z^{2} - 2 i z - 1 = 0

\frac{1 - 2 i}{4 + 2 i} = a + bi

x i n^{0} = 3 y^{2} + x - 1

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

What is the answer to -(2+3i)z-5+i=-z^2 ?
The answer to -(2+3i)z-5+i=-z^2 is z=-1+i,z=3+2i