Question

Rationalize the denominator of square root of negative 16 over open parentheses 1 plus i close parentheses plus open parentheses 6 plus 3 i.

Answer 1

$\bf \cfrac{\sqrt{-16}}{(1+i)+(6+3i)}\implies \cfrac{\sqrt{-1\cdot 16}}{1+i+6+3i}\implies \cfrac{\sqrt{-1}\cdot \sqrt{16}}{7+4i} \\\\\\ \cfrac{i\cdot \sqrt{4^2}}{7+4i}\implies \cfrac{4i}{7+4i}\impliedby \begin{array}{llll} \textit{now, we'll multiply by the}\\ \textit{conjugate of the denominator} \end{array}\\\\ -------------------------------$

$\bf \textit{and recall }\textit{difference of squares} \\ \quad \\ (a-b)(a+b) = a^2-b^2\qquad \qquad a^2-b^2 = (a-b)(a+b)\\\\ \textit{also recall that }i^2=-1 \\\\ -------------------------------$

$\bf \cfrac{4i}{7+4i}\cdot \cfrac{7-4i}{7-4i}\implies \cfrac{4i(7-4i)}{(7+4i)(7-4i)}\implies \cfrac{28i-16i^2}{7^2-(4i)^2} \\\\\\ \cfrac{28i-16(-1)}{49-(4^2i^2)}\implies \cfrac{28i+16}{49-[16(-1)]}\implies \cfrac{16+28i}{49+16}\implies \cfrac{16+28i}{65} \\\\\\ \cfrac{16}{65}+\cfrac{28i}{65}$