Question

2/3-i
how would I solve this?

Answer 1

to simplify or rationalize the denominator, you multiply both top and bottom by the conjugate of the denominator
so \(\bf \cfrac{2}{3-i}\\
\textit{let us use the conjugate }3+i\textit{ of the denominator}\\
\cfrac{2}{3-i}\times \cfrac{3+i}{3+i}\implies \cfrac{2(3+i)}{(3-i)(3+i)}\\ \quad \\
\textit{keep in mind that } (a-b)(a+b) = (a^2-b^2)\\ \quad \\
\cfrac{2(3+i)}{(3-i)(3+i)}\implies \cfrac{2(3+i)}{3-i^2}\)

Answer 2

Thanks so much!

Answer 3

wops, shooot... I have one mistake... one sec

Answer 4

\(\bf \cfrac{2(3+i)}{(3-i)(3+i)}\implies \cfrac{2(3+i)}{3^2-i^2}\)

Answer 5

when expanding the bottom, keep in mind that \(\bf i^2 \implies \sqrt{-1} \times \sqrt{-1} \implies \sqrt{(-1)^2}\implies -1\)

Answer 6

Thanks for your help!

Answer 7

yw

Answer 8

you have to do foil to the denominator right?

Answer 9

no, notice \(\bf i^2 \implies \sqrt{-1} \times \sqrt{-1} \implies \sqrt{(-1)^2}\implies -1\)

that means that \(\bf i^2 = -1\)
so
\(\bf \cfrac{2(3+i)}{3^2-i^2}\implies \cfrac{6+2i}{9-(-1)} \implies \cfrac{6+2i}{9+1}\)