Quadratic Equation/Examples/z^2 + (2i-3)z + 5-i = 0
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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: $32$