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5+3i , find the inverse multiplication

User German
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\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}
User Andorbal
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