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

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

has the solutions:

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

Proof

$\ds z^2 + \paren {i - 2} z + \paren {3 - i}$

$=$

$\ds 0$

$\ds \leadsto \ \ $

$\ds z$

$=$

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

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

$\ds $

$=$

$\ds \dfrac {2 - i \pm \sqrt {4 + i^2 - 4 i - 12 + 4 i} } 2$

simplifying

$\ds $

$=$

$\ds \dfrac {2 - i \pm \sqrt {-9} } 2$

simplifying

$\ds $

$=$

$\ds \dfrac {2 - i \pm 3 i} 2$

$\ds $

$=$

$\ds \begin{cases} 1 - 2 i \\ 1 + i \end{cases}$

$\blacksquare$

1981: Murray R. Spiegel: Theory and Problems of Complex Variables (SI ed.) ... (previous) ... (next): $1$: Complex Numbers: Supplementary Problems: Polynomial Equations: $100 \ \text{(b)}$