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

The answer to z^2+(i-2)z+(3+i)=0 is z=0.38339…+1.12178…i,z=1.61660…-2.12178…i

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z^2+(i-2)z+(3+i)=0

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

z = 0.38339 \dots + 1.12178 \dots i, z = 1.61660 \dots - 2.12178 \dots i

Solution steps

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

Substitute

z = x + y i

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} - 2 x - y + 3) + i (1 + x - 2 y + 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 + 3 = 0 1 + x - 2 y + 2 x y = 0]

[x^{2} - y^{2} - 2 x - y + 3 = 0 1 + x - 2 y + 2 x y = 0] : (x = 0.38339 \dots, x = 1.61660 \dots, y = 1.12178 \dots y = - 2.12178 \dots)

Substitute back

z = x + y i

z = 0.38339 \dots + 1.12178 \dots i, z = 1.61660 \dots - 2.12178 \dots i

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

z^{2} = 4 (- 1 - i 3)

i (t) = 4 t^{2} - 48 t + 154

i (t) = 8 t^{2} - 4 t A

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

What is the answer to z^2+(i-2)z+(3+i)=0 ?
The answer to z^2+(i-2)z+(3+i)=0 is z=0.38339…+1.12178…i,z=1.61660…-2.12178…i