Quadratic Equation/Examples/z^2 - 4z + 5 = 0

Example of Quadratic Equation
Let $b, c \in \R$ be real.

Let the quadratic equation:
 * $z^2 + b z + c = 0$

have the root:
 * $z = 2 + i$

Then the other root is:
 * $z = 2 - i$

and it follows that:

Proof
We are given that one of the roots of $z^2 + b z + c$ is $2 + i$.

It follows from Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs that the other root is $2 - i$.

From Viète's Formulas, it follows that:

Hence the result.