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:

\(\ds b\)

\(=\)

\(\ds -4\)

\(\ds c\)

\(=\)

\(\ds 5\)

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:

\(\ds b\)

\(=\)

\(\ds -\paren {\paren {2 + i} + \paren {2 - i} }\)

\(\ds \)

\(=\)

\(\ds -4\)

\(\ds c\)

\(=\)

\(\ds \paren {2 + i} \paren {2 - i}\)

\(\ds \)

\(=\)

\(\ds 4 - i^2\)

Difference of Two Squares

\(\ds \)

\(=\)

\(\ds 5\)

Hence the result.

$\blacksquare$

Sources

• 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 1$. Algebraic Theory of Complex Numbers: Exercise $8$