Question

5+3i , find the inverse multiplication

Answer 1

$\bf k\cdot (5+3i)=1\implies k=\cfrac{1}{5+3i}\impliedby \textit{multiplicative inverse} \\\\ \textit{now, the denominator, has a radical}\\ \textit{thus we have to rationalize it}\\ \textit{we'll use the "conjugate" of 5+3i, which is}\\ \textit{her sister, 5-3i} \\\\ \cfrac{1}{5+3i}\cdot \cfrac{5-3i}{5-3i}\implies \cfrac{5-3i}{(5+3i)(5-3i)}\\\\ -----------------------------\\\\ \textit{now, recall your }\textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)$

$\bf -----------------------------\\\\ thus \\\\ \cfrac{5-3i}{(5+3i)(5-3i)}\implies \cfrac{5-3i}{(5)^2-(3i)^2} \\\\\\ \cfrac{5-3i}{25-(3^2i^2)}\impliedby \textit{recall, }i^2=-1\qquad thus \\\\\\ \cfrac{5-3i}{25-(9\cdot -1)}\implies \cfrac{5-3i}{25+9}\implies \cfrac{5-3i}{34} \\\\\\ \textit{now, distributing the denominator, we get} \\\\ \cfrac{5}{34}-\cfrac{3i}{34}$