What is the answer to z^2+(2i+7)z+6+3i=0 ?

The answer to z^2+(2i+7)z+6+3i=0 is z=-1.06584…-0.17835…i,z=-5.93415…-1.82164…i

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z^2+(2i+7)z+6+3i=0

z^{2} + (2 i + 7) z + 6 + 3 i = 0

z = - 1.06584 \dots - 0.17835 \dots i, z = - 5.93415 \dots - 1.82164 \dots i

Solution steps

z^{2} + (2 i + 7) z + 6 + 3 i = 0

Substitute

z = x + y i

(x + y i)^{2} + (2 i + 7) (x + y i) + 6 + 3 i = 0

Expand

(x + y i)^{2} + (2 i + 7) (x + y i) + 6 + 3 i : (x^{2} - y^{2} + 7 x - 2 y + 6) + i (3 + 2 x + 7 y + 2 x y)

(x^{2} - y^{2} + 7 x - 2 y + 6) + i (3 + 2 x + 7 y + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} + 7 x - 2 y + 6) + i (3 + 2 x + 7 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} + 7 x - 2 y + 6 = 0 3 + 2 x + 7 y + 2 x y = 0]

[x^{2} - y^{2} + 7 x - 2 y + 6 = 0 3 + 2 x + 7 y + 2 x y = 0] : (x = - 1.06584 \dots, x = - 5.93415 \dots, y = - 0.17835 \dots y = - 1.82164 \dots)

Substitute back

z = x + y i

z = - 1.06584 \dots - 0.17835 \dots i, z = - 5.93415 \dots - 1.82164 \dots i

y, e = \frac{sin ( x )}{x e n y}

\frac{6 - 5 i}{2 + 3 i} = a + bi

x^{2} - y^{2} - 2 y x i = 4 x + 4

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

\frac{3 + ai}{5 + bi} = 2 + 7 i

What is the answer to z^2+(2i+7)z+6+3i=0 ?
The answer to z^2+(2i+7)z+6+3i=0 is z=-1.06584…-0.17835…i,z=-5.93415…-1.82164…i