If $g (x) = x^{4} - x^{3} \div x^{2} + 3 x - 5$ , find $g (2 + 3 i)$ ?

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Solution

$g (x) = \frac{x^{4} - x^{3}}{x^{2} + 3 x - 5} = \frac{x^{3} (x - 1)}{x^{2} + 3 x - 5}$
${(2 + 3 i)}^{3} = 8 - 27 i + 36 i - 54 = - 46 + 9 i$
${(2 + 3 i)}^{2} = 4 - 9 + 12 i = - 5 + 12 i$
Now, $g (2 + 3 i) = \frac{{(2 + 3 i)}^{3} ((2 + 3 i) - 1)}{{(2 + 3 i)}^{2} + 3 (2 + 3 i) - 5}$
$= \frac{(- 46 + 9 i) (1 + 3 i)}{(- 5 + 12 i) + (6 + 9 i) - 5} = \frac{- 46 - 138 i + 9 i - 27}{- 4 + 21 i}$
$= \frac{- 73 - 129 i}{- 4 + 21 i}$

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