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

The answer to z^2+(2i-3)z+5-1=0 is z=0.68768…+0.84657…i,z=2.31231…-2.84657…i

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z^2+(2i-3)z+5-1=0

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

z = 0.68768 \dots + 0.84657 \dots i, z = 2.31231 \dots - 2.84657 \dots i

Solution steps

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

Substitute

z = x + y i

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

Expand

(x + y i)^{2} + (2 i - 3) (x + y i) + 5 - 1 : (x^{2} - 3 x + 4 - y^{2} - 2 y) + i (2 x - 3 y + 2 x y)

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

Rewrite

0

in standard complex form:

0 + 0 i

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

[x^{2} - 3 x + 4 - y^{2} - 2 y = 0 2 x - 3 y + 2 x y = 0] : (x = 0.68768 \dots, x = 2.31231 \dots, y = 0.84657 \dots y = - 2.84657 \dots)

Substitute back

z = x + y i

z = 0.68768 \dots + 0.84657 \dots i, z = 2.31231 \dots - 2.84657 \dots i

s i f (x) = \frac{5 x ^{2}}{x}, f (0)

z^{2} + 4 z i - 4 = 0

\frac{( a + bi ) - 3}{3 - ( a + bi )} = 3

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

2 i x_{2} = - x_{3}

What is the answer to z^2+(2i-3)z+5-1=0 ?
The answer to z^2+(2i-3)z+5-1=0 is z=0.68768…+0.84657…i,z=2.31231…-2.84657…i