What is the answer to (1+i)z^2+2iz+(-5+5i)=0 ?

The answer to (1+i)z^2+2iz+(-5+5i)=0 is z=1-2i,z=-2+i

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(1+i)z^2+2iz+(-5+5i)=0

(1 + i) z^{2} + 2 i z + (- 5 + 5 i) = 0

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

Solution steps

(1 + i) z^{2} + 2 i z + (- 5 + 5 i) = 0

Substitute

z = x + y i

(1 + i) (x + y i)^{2} + 2 i (x + y i) - 5 + 5 i = 0

Expand

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

(x^{2} - y^{2} - 2 x y - 2 y - 5) + i (5 + x^{2} - y^{2} + 2 x + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} - 2 x y - 2 y - 5) + i (5 + x^{2} - y^{2} + 2 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} - 2 x y - 2 y - 5 = 0 5 + x^{2} - y^{2} + 2 x + 2 x y = 0]

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

Substitute back

z = x + y i

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

7 - 4 y i = 9 x + 3 i

t, \frac{L d i}{d t} + R i = E

15 - 3 y i = \frac{1}{2} x + 2 i

(9 x + 8 y) + i (12 x - 6 y) = 10 + 10 i

z^{2} = - 1 + 3 i

What is the answer to (1+i)z^2+2iz+(-5+5i)=0 ?
The answer to (1+i)z^2+2iz+(-5+5i)=0 is z=1-2i,z=-2+i