What is the answer to z^2-(3+4i)z+(1-9i)=0 ?

The answer to z^2-(3+4i)z+(1-9i)=0 is z=-1-i,z=4+5i

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z^2-(3+4i)z+(1-9i)=0

z^{2} - (3 + 4 i) z + (1 - 9 i) = 0

z = - 1 - i, z = 4 + 5 i

Solution steps

z^{2} - (3 + 4 i) z + (1 - 9 i) = 0

Substitute

z = x + y i

(x + y i)^{2} - (3 + 4 i) (x + y i) + 1 - 9 i = 0

Expand

(x + y i)^{2} - (3 + 4 i) (x + y i) + 1 - 9 i : (x^{2} - y^{2} - 3 x + 4 y + 1) + i (- 9 - 3 y - 4 x + 2 x y)

(x^{2} - y^{2} - 3 x + 4 y + 1) + i (- 9 - 3 y - 4 x + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} - 3 x + 4 y + 1) + i (- 9 - 3 y - 4 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 x + 4 y + 1 = 0 - 9 - 3 y - 4 x + 2 x y = 0]

[x^{2} - y^{2} - 3 x + 4 y + 1 = 0 - 9 - 3 y - 4 x + 2 x y = 0] : (x = - 1, x = 4, y = - 1 y = 5)

Substitute back

z = x + y i

z = - 1 - i, z = 4 + 5 i

C = M (1 + t i)

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

x y i - 4 (x + y^{6}) j = 0

\frac{m}{1 + i} + n = \frac{- m - 1}{i}

z^{2} + 2 i z^{2} + 3 i z = 0

What is the answer to z^2-(3+4i)z+(1-9i)=0 ?
The answer to z^2-(3+4i)z+(1-9i)=0 is z=-1-i,z=4+5i