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

6 plus 3 i.

Mathematics

1 answer:

emmasim [6.3K]3 years ago

5 0

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

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

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

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