[tex]\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)\\\\
[/tex]
[tex]\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}[/tex]
