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z^2-(2-i)z+3-i=0 equation
The teacher will be very surprised to see your correct solution 😉
The solution
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[src]
2 z - (2 - I)*z + 3 - I = 0
$$\left(\left(z^{2} - z \left(2 - i\right)\right) + 3\right) - i = 0$$
Detail solution
Expand the expression in the equation
$$\left(\left(z^{2} - z \left(2 - i\right)\right) + 3\right) - i = 0$$
We get the quadratic equation
$$z^{2} - 2 z + i z + 3 - i = 0$$
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
$$a = 1$$
$$b = -2 + i$$
$$c = 3 - i$$
, then
The equation has two roots.
or
$$z_{1} = 1 - \frac{i}{2} + \frac{\sqrt{-12 + \left(-2 + i\right)^{2} + 4 i}}{2}$$
$$z_{2} = 1 - \frac{\sqrt{-12 + \left(-2 + i\right)^{2} + 4 i}}{2} - \frac{i}{2}$$
$$\left(\left(z^{2} - z \left(2 - i\right)\right) + 3\right) - i = 0$$
We get the quadratic equation
$$z^{2} - 2 z + i z + 3 - i = 0$$
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
$$a = 1$$
$$b = -2 + i$$
$$c = 3 - i$$
, then
D = b^2 - 4 * a * c =
(-2 + i)^2 - 4 * (1) * (3 - i) = -12 + (-2 + i)^2 + 4*i
The equation has two roots.
z1 = (-b + sqrt(D)) / (2*a)
z2 = (-b - sqrt(D)) / (2*a)
or
$$z_{1} = 1 - \frac{i}{2} + \frac{\sqrt{-12 + \left(-2 + i\right)^{2} + 4 i}}{2}$$
$$z_{2} = 1 - \frac{\sqrt{-12 + \left(-2 + i\right)^{2} + 4 i}}{2} - \frac{i}{2}$$
Vieta's Theorem
it is reduced quadratic equation
$$p z + q + z^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2 + i$$
$$q = \frac{c}{a}$$
$$q = 3 - i$$
Vieta Formulas
$$z_{1} + z_{2} = - p$$
$$z_{1} z_{2} = q$$
$$z_{1} + z_{2} = 2 - i$$
$$z_{1} z_{2} = 3 - i$$
$$p z + q + z^{2} = 0$$
where
$$p = \frac{b}{a}$$
$$p = -2 + i$$
$$q = \frac{c}{a}$$
$$q = 3 - i$$
Vieta Formulas
$$z_{1} + z_{2} = - p$$
$$z_{1} z_{2} = q$$
$$z_{1} + z_{2} = 2 - i$$
$$z_{1} z_{2} = 3 - i$$
The graph
Sum and product of roots
[src]
sum
1 - 2*I + 1 + I
$$\left(1 - 2 i\right) + \left(1 + i\right)$$
=
2 - I
$$2 - i$$
product
(1 - 2*I)*(1 + I)
$$\left(1 - 2 i\right) \left(1 + i\right)$$
=
3 - I
$$3 - i$$
3 - i