What is the answer to z^2+(2i-3)z+(5-i)=0 ?

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

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z^2+(2i-3)z+(5-i)=0

z^{2} + (2 i - 3) z + (5 - i) = 0

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

Solution steps

z^{2} + (2 i - 3) z + (5 - i) = 0

Substitute

z = x + y i

(x + y i)^{2} + (2 i - 3) (x + y i) + 5 - i = 0

Expand

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

(x^{2} - y^{2} - 3 x - 2 y + 5) + i (- 1 + 2 x - 3 y + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} - 3 x - 2 y + 5) + i (- 1 + 2 x - 3 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} - 3 x - 2 y + 5 = 0 - 1 + 2 x - 3 y + 2 x y = 0]

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

Substitute back

z = x + y i

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

a, \frac{1}{a + bi} = c + d i

5 + (x^{2} - 6 x + 9) i = 0

3 z^{2} - (5 - 4 i) z + (2 - 3 i) = 0

(a - bi) ((a + bi) + 1) = 11 + 3 i

x - i y = 2 + 5 i

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