Quadratic Equation/Examples/z^2 - (3+i)z + 4+3i = 0
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Example of Quadratic Equation
The quadratic equation in $\C$:
- $z^2 - \paren {3 + i} z + 4 + 3 i = 0$
has the solutions:
- $z = \begin{cases} 1 + 2 i \\ 2 - i \end{cases}$
Proof
From the Quadratic Formula:
\(\ds z^2 - \paren {3 + i} z + 4 + 3 i\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds \dfrac {-\paren {-\paren {3 + i} } \pm \sqrt {\paren {-\paren {3 + i} }^2 - 4 \times 1 \times \paren {4 + 3 i} } } {2 \times 1}\) | Quadratic Formula: $a = 1$, $b = -\paren {3 + i}$, $c = 4 + 3 i$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {3 + i} 2 \pm \dfrac {\sqrt {\paren {8 + 6 i} - \paren {16 + 12 i} } } 2\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {3 + i} 2 \pm \dfrac {\sqrt {-8 - 6 i} } 2\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {3 + i} 2 \pm \dfrac {\sqrt {\paren {-1} \paren {8 + 6 i} } } 2\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {3 + i} 2 \pm \dfrac {i \sqrt {\paren {8 + 6 i} } } 2\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {3 + i} 2 \pm \dfrac {i \paren {3 + i} } 2\) | from above: we have established that $\paren {\pm \paren {3 + i} }^2 = \paren {8 + 6 i}$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {3 + i} 2 \pm \dfrac {3 i + i^2} 2\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {3 + i} 2 \pm \dfrac {-1 + 3 i} 2\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \begin{cases} 1 + 2 i \\ 2 - i \end{cases}\) | simplifying |
$\blacksquare$
Sources
- 1960: Walter Ledermann: Complex Numbers ... (previous) ... (next): $\S 1$. Algebraic Theory of Complex Numbers: Exercise $7$