Question

Advance math
if (x + jy)^2=3 +j8,find x and y

Answer 1

what is j?

Answer 2

i think that j is i
then,
x^2 +2xy j + (jy)^2 = 3 +j8
x^2-y^2 +2xyj = 3+j8
then x^2-y^2 = 3 and 2xyj = j8
now use substitution method to find x and y

Answer 3

are you kidding? you have to solve
\[x^2-y^2=3, 2xy=8\]?? good luck!

Answer 4

huh??i don't understand satellite..the problem is finding x and y..How should i do it??

Answer 5

Hey saruz how did you do it.can you explain how did you came up with that answer?please.

Answer 6

assuming
\[j^2=-1\] you get
\[(x+yj)^2=x^2+2xyj+yj^2=x^2-y^2+2xyj\]

Answer 7

and this means that
\[x^2-y^2=3\] and
\[2xy=8\] but i have no idea how you are supposed to solve that one. are you sure this is the problem?

Answer 8

you can see here how impossible this is to solve
http://www.wolframalpha.com/input/?i=x^2-y^2%3D3+and+2xy%3D8

Answer 9

by using the theorem of complex numbers..how to solve x and y?

Answer 10

on the other hand, we are both making the assumption that
\[j^2=-1\] which is a common notation. maybe it is something else

Answer 11

ok first of all usually when you write a complex number in standard for
\[x+yj\] x and y are real. but in this case they are not. as you can see from worfram
also as you can see from wolfram there is no easy ( or even reasonable) way to solve this problem, so i guess i am stumped. look at the solution and you will see how odd it is.

Answer 12

satellite I have another problem here can you help me answer this.
if (x+jy)^3=2-j4, find x and y

Answer 13

\[x^2-y^2=3\quad,\quad xy=4\]\[x^2y^2=x^2(x^2-3)=16\]\[x^4-3x^2-16=0\]\[x^2=\frac{3+\sqrt{73}}{2}\quad,\quad y^2=x^2-3=\frac{-3+\sqrt{73}}{2}\]\[(x,y)=\left\{\left(\sqrt{\frac{3+\sqrt{73}}{2}},\sqrt{\frac{-3+\sqrt{73}}{2}}\right),\left(-\sqrt{\frac{3+\sqrt{73}}{2}},-\sqrt{\frac{-3+\sqrt{73}}{2}}\right)\right\}\]