Question

write (7-i)/(3-2i) in standard form for a complex number

Answer 1

do you know what the conjugate is?

Answer 2

nope

Answer 3

(x+y) and (x-y) are conjugates. they're the same except for the sign. the deal with conjugates is that when you multiply them, you don't get a middle term.

(x+y)(x-y) = x^2 + xy - xy - y^2 = x^2 - y^2

to divide by a complex number, you simply multiply the top and bottom by the conjugate of the bottom. this get's rid of the i and you're left with a real number on the bottom.

Answer 4

so how do i begin to solve this?

Answer 5

what number are you dividing by?

Answer 6

what is its conjugate?

Answer 7

multiply top and bottom by the conjugate. give it a shot and if you get stuck, i'm here.

Answer 8

\[\frac{ 7-i }{ 3-2i }\]
so that is how it is set up...now what? lol sorry !

Answer 9

what is on the bottom? what is its conjugate?

Answer 10

the 3-2i?

Answer 11

yes... now what is its conjugate?

Answer 12

so multiply the top and bottom by the 3-2i

Answer 13

no... by the conjugate of that.

Answer 14

x-y?

Answer 15

change the sign inbetween... that's all you do!

Answer 16

so the conjugate of 3 - 2i is just: 3 change the sign 2i

Answer 17

what should the sign be?

Answer 18

+

Answer 19

yeah!!!

so multiply top and bottom by 3 + 2i...

Answer 20

\[\frac{ 7-i }{ 3-2i }\cdot \frac{ 3+2i}{ 3+2i }=\frac{ (7-i)(3+2i)}{ 3^{2}+2^{2} }\]
you can finish it off...

Answer 21

so i got.. \[\frac{ 21-2i ^{2} }{ 9-4i ^{2}}\] ??

Answer 22

almost...
\[\frac{ (7-i)(3+2i) }{ 3^2-4i^2 }=\frac{ (21-2i^2+14i-3i) }{ 9-4(-1) }=\frac{ (21-2(-1)+14i-3i) }{ 9+4 }\]
can you get it from here?

Answer 23

\[\frac{ 12 }{ 13 }\] ?

Answer 24

\[\frac{ (23+11i) }{ 13 }=\frac{ 23 }{ 13 }+\frac{11}{ 13 }i\]
sorry... got to run!
good luck!

Answer 25

ok thanks!!