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`


`\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}`