Question

Simplify.

2 + 4i/3i

Answer 1

simplify my eye! it means "divide and write an standard form
there is no such mathematical operation as "simplify"

Answer 2

i assume it is
\[\frac{2+4i}{3i}\] right?

Answer 3

yeup!

Answer 4

multiply top and bottom by \(i\)

Answer 5

then there will be no \(i\) in the denominator
let me know what you get

Answer 6

\[\frac{ 2i-4 }{-3 }\]

Answer 7

@satellite73

Answer 8

ok good one step away from writing in standard form

Answer 9

\[\frac{2i-4}{-3}=\frac{4}{3}-\frac{2}{3}i\]

Answer 10

good with that?

Answer 11

would I simplify this the same way? \[\frac{ 3+2i }{ 4+i }\]

Answer 12

no

Answer 13

if you multiply top and bottom by \(i\) you will still have an \(i\) in the denominator, it will just change places

Answer 14

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

Answer 15

When I used FOIL to multiply it I got \[\frac{ 12-3i+8i-2i ^{2} }{ 16-4i-i ^{2}+4i }\]

Answer 16

ok don't forget that \((a+bi)(a-bi)=a^2+b^2\) so your denominator should be \(4^2+1^2=17\)

Answer 17

which is in fact what you have, just too much work

Answer 18

oh OK so now I got \[\frac{ 14+5i }{ 17 }\] is that it? or can we do more to that?

Answer 19

if you want it in standard for you write it as
\[\frac{14}{17}+\frac{5}{17}i\]

Answer 20

"standard forM"

Answer 21

alright I got it :)