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z^2-(2+i)z+2i=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
Detail solution
Expand the expression in the equation
$$\left(z^{2} - z \left(2 + i\right) + 2 i\right) + 0 = 0$$
We get the quadratic equation
$$z^{2} - 2 z - i z + 2 i = 0$$
This equation is of the form
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$z_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$z_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -2 - i$$
$$c = 2 i$$
, then
The equation has two roots.
or
$$z_{1} = 1 + \frac{\sqrt{- 8 i + \left(-2 - i\right)^{2}}}{2} + \frac{i}{2}$$
Simplify
$$z_{2} = 1 - \frac{\sqrt{- 8 i + \left(-2 - i\right)^{2}}}{2} + \frac{i}{2}$$
Simplify
$$\left(z^{2} - z \left(2 + i\right) + 2 i\right) + 0 = 0$$
We get the quadratic equation
$$z^{2} - 2 z - i z + 2 i = 0$$
This equation is of the form
a*z^2 + b*z + c = 0
A quadratic equation can be solved
using the discriminant.
The roots of the quadratic equation:
$$z_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$z_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where D = b^2 - 4*a*c - it is the discriminant.
Because
$$a = 1$$
$$b = -2 - i$$
$$c = 2 i$$
, then
D = b^2 - 4 * a * c =
(-2 - i)^2 - 4 * (1) * (2*i) = (-2 - i)^2 - 8*i
The equation has two roots.
z1 = (-b + sqrt(D)) / (2*a)
z2 = (-b - sqrt(D)) / (2*a)
or
$$z_{1} = 1 + \frac{\sqrt{- 8 i + \left(-2 - i\right)^{2}}}{2} + \frac{i}{2}$$
Simplify
$$z_{2} = 1 - \frac{\sqrt{- 8 i + \left(-2 - i\right)^{2}}}{2} + \frac{i}{2}$$
Simplify
Vieta's Theorem
it is reduced quadratic equation
$$p z + q + z^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2 - i$$
$$q = \frac{c}{a}$$
$$q = 2 i$$
Vieta Formulas
$$z_{1} + z_{2} = - p$$
$$z_{1} z_{2} = q$$
$$z_{1} + z_{2} = 2 + i$$
$$z_{1} z_{2} = 2 i$$
$$p z + q + z^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2 - i$$
$$q = \frac{c}{a}$$
$$q = 2 i$$
Vieta Formulas
$$z_{1} + z_{2} = - p$$
$$z_{1} z_{2} = q$$
$$z_{1} + z_{2} = 2 + i$$
$$z_{1} z_{2} = 2 i$$
The graph