anonymous
  • anonymous
write (7-i)/(3-2i) in standard form for a complex number
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
do you know what the conjugate is?
anonymous
  • anonymous
nope
anonymous
  • anonymous
(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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
so how do i begin to solve this?
anonymous
  • anonymous
what number are you dividing by?
anonymous
  • anonymous
what is its conjugate?
anonymous
  • anonymous
multiply top and bottom by the conjugate. give it a shot and if you get stuck, i'm here.
anonymous
  • anonymous
\[\frac{ 7-i }{ 3-2i }\] so that is how it is set up...now what? lol sorry !
anonymous
  • anonymous
what is on the bottom? what is its conjugate?
anonymous
  • anonymous
the 3-2i?
anonymous
  • anonymous
yes... now what is its conjugate?
anonymous
  • anonymous
so multiply the top and bottom by the 3-2i
anonymous
  • anonymous
no... by the conjugate of that.
anonymous
  • anonymous
x-y?
anonymous
  • anonymous
change the sign inbetween... that's all you do!
anonymous
  • anonymous
so the conjugate of 3 - 2i is just: 3 change the sign 2i
anonymous
  • anonymous
what should the sign be?
anonymous
  • anonymous
+
anonymous
  • anonymous
yeah!!! so multiply top and bottom by 3 + 2i...
anonymous
  • anonymous
\[\frac{ 7-i }{ 3-2i }\cdot \frac{ 3+2i}{ 3+2i }=\frac{ (7-i)(3+2i)}{ 3^{2}+2^{2} }\] you can finish it off...
anonymous
  • anonymous
so i got.. \[\frac{ 21-2i ^{2} }{ 9-4i ^{2}}\] ??
anonymous
  • anonymous
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?
anonymous
  • anonymous
\[\frac{ 12 }{ 13 }\] ?
anonymous
  • anonymous
\[\frac{ (23+11i) }{ 13 }=\frac{ 23 }{ 13 }+\frac{11}{ 13 }i\] sorry... got to run! good luck!
anonymous
  • anonymous
ok thanks!!

Looking for something else?

Not the answer you are looking for? Search for more explanations.