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

The answer to z^2+(3-i)z+(2+i)=0 is z=-0.38196…-0.61803…i,z=-2.61803…+1.61803…i

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

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

z = - 0.38196 \dots - 0.61803 \dots i, z = - 2.61803 \dots + 1.61803 \dots i

Solution steps

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

Substitute

z = x + y i

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

[x^{2} - y^{2} + 3 x + y + 2 = 0 1 - x + 3 y + 2 x y = 0] : (x = - 0.38196 \dots, x = - 2.61803 \dots, y = - 0.61803 \dots y = 1.61803 \dots)

Substitute back

z = x + y i

z = - 0.38196 \dots - 0.61803 \dots i, z = - 2.61803 \dots + 1.61803 \dots i

i \frac{1}{16} x + 6 y = 7

15 - 28 i = 3 l + (4 m) i

x^{2} = 1184 - 125.6637 i

(x + 2 i) (y + 2 i) = \frac{16 i}{1 + i} - 9

x, z = \frac{x + 4 i}{1 + x i}

What is the answer to z^2+(3-i)z+(2+i)=0 ?
The answer to z^2+(3-i)z+(2+i)=0 is z=-0.38196…-0.61803…i,z=-2.61803…+1.61803…i