Question

Please help me simplify

Answer 1

\[\frac{ 3-i }{ 2+5i }\]

Answer 2

god help me they call everything "simplify"
what they mean is "write in standard from as \(a+bi\)"
tell your math teacher !

Answer 3

Now everything makes sense

Answer 4

\[\frac{ 3-i }{ 2+5i}\times \frac{2-5i}{2-5i}\] is a start

Answer 5

the conjugate right?... I don't know if thats how you spell it

Answer 6

to eliminate the i in the bottom

Answer 7

i.e. multiply 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 8

on your case you will have
\[\frac{(3-i)(2-5i)}{2^2+5^2}\] all the work is now in the numerator

Answer 9

yeah that is how you spell it too

Answer 10

you got the top?

Answer 11

yeah that would be \[\frac{ 4-10i }{ 4-25i^2 }\]

Answer 12

right? @satellite73

Answer 13

My bad, I mean \[\frac{ 1-17i }{ 4-25i^2 }\]

Answer 14

right? @satellite73

Answer 15

\[(a+bi)(a-bi)=a^2+b^2\] so
\[(2+5i)(2-5i)=2^2+5^2=4+25=29\]

Answer 16

dont mess around with any \(i\) stuff when you multiply by the conjugate

Answer 17

all right so it would be \[\frac{ 1 }{ 29 }-\frac{ 17i }{ 29 }\]

Answer 18

right? @misty1212