What is the answer to (1+i3)z^2+(5+i6)z+7-i6=0 ?

The answer to (1+i3)z^2+(5+i6)z+7-i6=0 is

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(1+i3)z^2+(5+i6)z+7-i6=0

(1 + i 3) z^{2} + (5 + i 6) z + 7 - i 6 = 0

z = \frac{- 5.46069 \dots + 1.2 E - 14}{2} - 0.07678 \dots i, z = \frac{- 5.46069 \dots - 1.2 E - 14}{2} - 0.07678 \dots i, z = \frac{0.86069 \dots + 1.11 E - 14}{2} + 0.97678 \dots i, z = \frac{0.86069 \dots - 1.11 E - 14}{2} + 0.97678 \dots i

Solution steps

(1 + i 3) z^{2} + (5 + i 6) z + 7 - i 6 = 0

Substitute

z = x + y i

(1 + i 3) (x + y i)^{2} + (5 + i 6) (x + y i) + 7 - i 6 = 0

Expand

(1 + i 3) (x + y i)^{2} + (5 + i 6) (x + y i) + 7 - i 6 : (x^{2} - y^{2} - 6 x y + 5 x - 6 y + 7) + i (- 6 + 3 x^{2} - 3 y^{2} + 5 y + 6 x + 2 x y)

(x^{2} - y^{2} - 6 x y + 5 x - 6 y + 7) + i (- 6 + 3 x^{2} - 3 y^{2} + 5 y + 6 x + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} - 6 x y + 5 x - 6 y + 7) + i (- 6 + 3 x^{2} - 3 y^{2} + 5 y + 6 x + 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} - 6 x y + 5 x - 6 y + 7 = 0 - 6 + 3 x^{2} - 3 y^{2} + 5 y + 6 x + 2 x y = 0]

[x^{2} - y^{2} - 6 x y + 5 x - 6 y + 7 = 0 - 6 + 3 x^{2} - 3 y^{2} + 5 y + 6 x + 2 x y = 0] : x = \frac{- 5.46069 \dots + 1.2 E - 14}{2}, x = \frac{- 5.46069 \dots - 1.2 E - 14}{2}, x = \frac{0.86069 \dots + 1.11 E - 14}{2}, x = \frac{0.86069 \dots - 1.11 E - 14}{2}, x = \frac{- 5.46069 \dots + 1.2 E - 14}{2}, x = \frac{- 5.46069 \dots - 1.2 E - 14}{2}, x = \frac{0.86069 \dots + 1.11 E - 14}{2}, x = \frac{0.86069 \dots - 1.11 E - 14}{2}, y = - 0.07678 \dots y = - 0.07678 \dots y = 0.97678 \dots y = 0.97678 \dots y = - 0.07678 \dots y = - 0.07678 \dots y = 0.97678 \dots y = 0.97678 \dots

Substitute back

z = x + y i

z = \frac{- 5.46069 \dots + 1.2 E - 14}{2} - 0.07678 \dots i, z = \frac{- 5.46069 \dots - 1.2 E - 14}{2} - 0.07678 \dots i, z = \frac{0.86069 \dots + 1.11 E - 14}{2} + 0.97678 \dots i, z = \frac{0.86069 \dots - 1.11 E - 14}{2} + 0.97678 \dots i

a^{2} + b^{2} - (a + bi) = 1

i = \frac{v}{r}

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

+ 2 i c^{2} = - 8 + 3 i, c^{1} = 5

x + i y = 3 i - i x

What is the answer to (1+i3)z^2+(5+i6)z+7-i6=0 ?
The answer to (1+i3)z^2+(5+i6)z+7-i6=0 is