Question

Rationalize the denominator

Answer 1

\[\frac{ {\sqrt{4}} }{ (7-3i+(2+5i) }\]

Answer 2

\(\bf \cfrac{ {\sqrt{4}} }{ (7-3i)+(2+5i) }\quad ?\)

Answer 3

yes

Answer 4

well... if you remove the parentheses for the denominator... what would it end up as?

Answer 5

umm 9+2i?

Answer 6

yeap... so to "rationalize it" is to get rid of that pesky "i" at the bottom
you'd do that by simply multiplying top and bottom by the
denominator's conjugate

9+2i conjugate -> 9-2i

so \(\bf \cfrac{ {\sqrt{4}} }{ (7-3i)+(2+5i) }\implies \cfrac{ {\sqrt{4}} }{ 7-3i+2+5i }\implies \cfrac{ {\sqrt{4}} }{ 9+2i }
\\ \quad \\
\cfrac{ {\sqrt{4}} }{ 9+2i }\cdot \cfrac{ 9-2i }{ 9-2i }\implies \cfrac{ {\sqrt{4}}(9-2i) }{ (9+2i)(9-2i) }
\\ \quad \\
recall\implies {\color{brown}{ (a-b)(a+b) = a^2-b^2}}\qquad thus
\\ \quad \\
\cfrac{ {\sqrt{4}}(9-2i) }{ (9+2i)(9-2i) }\implies \cfrac{ {\sqrt{4}}(9-2i) }{ 9^2-(2i)^2 }\implies \cfrac{ {\sqrt{4}}(9-2i) }{ 9^2- 2^2 i^2 }
\\ \quad \\
recall\implies {\color{brown}{ i^2\to \sqrt{-1}\cdot \sqrt{-1}\to \sqrt{(-1)^2}\to -1}}\qquad thus
\\ \quad \\
\cfrac{ {\sqrt{4}}(9-2i) }{ 9^2- 2^2\cdot -1 }\implies \cfrac{ {\sqrt{4}}(9-2i) }{81+2}\)

Answer 7

is that it

Answer 8

the answer

Answer 9

actually \(\bf \cfrac{ {\sqrt{4}}(9-2i) }{ 9^2- 2^2\cdot -1 }\implies \cfrac{ {\sqrt{4}}(9-2i) }{81+4}\)

Answer 10

\(\bf \cfrac{ {\sqrt{4}}(9-2i) }{ 9^2- 2^2\cdot -1 }\implies \cfrac{ {\sqrt{4}}(9-2i) }{81+4}\implies \cfrac{9\sqrt{4}-2i\sqrt{4}}{85}
\\ \quad \\
notice\implies {\color{brown}{ \sqrt{4}\to \sqrt{2^2}\to 2}}
\\ \quad \\
\cfrac{9\sqrt{4}-2i\sqrt{4}}{85}\implies \cfrac{9\cdot 2-2i\cdot 2}{85}\implies \cfrac{18-4i}{85}\)