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

The answer to z^2-(5+2i)z+2=0 is z=0.35367…-0.16478…i,z=4.64632…+2.16478…i

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

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

z = 0.35367 \dots - 0.16478 \dots i, z = 4.64632 \dots + 2.16478 \dots i

Solution steps

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

Substitute

z = x + y i

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

[x^{2} - y^{2} - 5 x + 2 y + 2 = 0 - 2 x - 5 y + 2 x y = 0] : (x = 0.35367 \dots, x = 4.64632 \dots, y = - 0.16478 \dots y = 2.16478 \dots)

Substitute back

z = x + y i

z = 0.35367 \dots - 0.16478 \dots i, z = 4.64632 \dots + 2.16478 \dots i

x + i y - 1 + i = 3

\frac{( x + i y )}{x - i y} = - i

x^{2} - 1 = i

x^{2} - (7 + 3 i) x + (10 + 11 i) = 0

∣ a - 1 + 2 i ∣ = 3

What is the answer to z^2-(5+2i)z+2=0 ?
The answer to z^2-(5+2i)z+2=0 is z=0.35367…-0.16478…i,z=4.64632…+2.16478…i