How about we think about \[z\text{ }=\text{ }\surd 5\text{ }+\text{ }3i\]
\[{{\left| z \right|}^{2}}~=\text{ }{{\left( \surd 5 \right)}^{2}}~+\text{ }{{3}^{2}}~=\text{ }5\text{ }+\text{ }9\text{ }=\text{ }14\]
In this manner, the multiplicative converse of \[\surd 5\text{ }+\text{ }3i\]is given by \[z-1\]
