What is the answer to z^2-(6-2i)z+(11-10i)=0 ?

The answer to z^2-(6-2i)z+(11-10i)=0 is z=2-3i,z=4+i

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z^2-(6-2i)z+(11-10i)=0

z^{2} - (6 - 2 i) z + (11 - 10 i) = 0

z = 2 - 3 i, z = 4 + i

Solution steps

z^{2} - (6 - 2 i) z + (11 - 10 i) = 0

Substitute

z = x + y i

(x + y i)^{2} - (6 - 2 i) (x + y i) + 11 - 10 i = 0

Expand

(x + y i)^{2} - (6 - 2 i) (x + y i) + 11 - 10 i : (x^{2} - y^{2} - 6 x - 2 y + 11) + i (- 10 + 2 x - 6 y + 2 x y)

(x^{2} - y^{2} - 6 x - 2 y + 11) + i (- 10 + 2 x - 6 y + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} - 6 x - 2 y + 11) + i (- 10 + 2 x - 6 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} - 6 x - 2 y + 11 = 0 - 10 + 2 x - 6 y + 2 x y = 0]

[x^{2} - y^{2} - 6 x - 2 y + 11 = 0 - 10 + 2 x - 6 y + 2 x y = 0] : (x = 2, x = 4, y = - 3 y = 1)

Substitute back

z = x + y i

z = 2 - 3 i, z = 4 + i

i = 1.4 x 1 0^{- 4} + 4.363 x 1 0^{- 4} i

i w^{2} + 8 w - 7 i = 0

z \cdot (z + 1) = 5 + 5 i

\frac{( z + 1 )}{z - 1} = i

(2 + i)^{2} = c + d i

What is the answer to z^2-(6-2i)z+(11-10i)=0 ?
The answer to z^2-(6-2i)z+(11-10i)=0 is z=2-3i,z=4+i