Question

Write the expression in standard form.
3/(3-12i)

Select one:

a. (1/-17)+(4/17)i
b. (1/17)-(4/17)i
c. (1/-17)-(4/17)i
d. (1/17)+(4/17)i

Answer 1

\[\frac{3}{3-12i}\] multiply top and bottom by the conjugate of the denominator

Answer 2

the conjugate of \(a+bi\) is \(a-bi\) and this works because \[(a+bi)(a-bi)=a^2+b^2\] a real number

Answer 3

Thank you.

Answer 4

\[\frac{3}{3-12i}\times \frac{3+12i}{3-12i}=\frac{3(3+12i)}{3^2+12^2}\]

Answer 5

the numerator is what you get when you multiply
i would leave it in factored form first, because you are going to cancel the three

Answer 6

which, now that i look, you could have cancelled before starting \[\frac{3}{3-12i}=\frac{1}{1-4i}\] probably would have been easier to start there

Answer 7

oh , you are welcome

Answer 8

\(\color{blue}{\text{Originally Posted by}}\) satellite73
\[\frac{3}{3-12i}\times \frac{3+12i}{3-12i}=\frac{3(3+12i)}{3^2+12^2}\]
\(\color{blue}{\text{End of Quote}}\)
looks like there is a typo \[\frac{3}{3-12i}\times \frac{3+12i}{3\color{Red}{+}12i}\]
??