Question

hey can anyone help me with this problem? write the expression in standard form. 3 over 3-12i please help

Answer 1

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

Answer 2

\[\frac{3}{3-12i}\times \frac{3+12i}{3+12i}\]is a start

Answer 3

course we could have been smarter and cancelled a 3 at the beginning and started with \[\frac{1}{1-4i}\times \frac{1+4i}{1+4i}\] to make life easier

Answer 4

ok so I got 1 over 17- 4 over 17 i.. right?

Answer 5

ooh wait i am sorry hold on

Answer 6

i read your answer incorrectly
you meant
\[\frac{1}{17}-\frac{4}{17}i\] right ?

Answer 7

yes

Answer 8

almost

Answer 9

\[\frac{1}{1-4i}\times \frac{1+4i}{1+4i}=\frac{1+4i}{17}=\frac{1}{17}+\frac{4}{17}i\]

Answer 10

ooh because i^2 is equal to -1 correct? i see that on some other cites

Answer 11

yes that is the definition of \(i\)

Answer 12

ok i got it now thank you so much!