Quadratic Equation/Examples/z^2 + (2i-3)z + 5-i = 0

Example of Quadratic Equation

The quadratic equation in $\C$:

$z^2 + \paren {2 i - 3} z + 5 - i = 0$

has the solutions:

$z = \begin{cases} 2 - 3 i \\ 1 + i \end{cases}$

Proof

From the Quadratic Formula:

\(\ds z^2 + \paren {2 i - 3} z + 5 - i\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds z\)

\(=\)

\(\ds \dfrac {-\paren {2 i - 3} \pm \sqrt {\paren {2 i - 3}^2 - 4 \times 1 \times \paren {5 - i} } } {2 \times 1}\)

Quadratic Formula: $a = 1$, $b = 2 i - 3$, $c = 5 - i$

\(\ds \)

\(=\)

\(\ds \dfrac {3 - 2 i \pm \sqrt {-15 - 8 i} } 2\)

simplifying

\(\ds \)

\(=\)

\(\ds \dfrac {3 - 2 i \pm \paren {1 - 4 i} } 2\)

Square Root of $-15 - 8 i$

\(\ds \)

\(=\)

\(\ds \begin{cases} 2 - 3 i \\ 1 + i \end{cases}\)

simplifying

$\blacksquare$

Sources

• 1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Solved Problems: Polynomial Equations: $32$