Question

If (x+iy)^2=15+8i, and x and y are real #'s. and x>0, what is x-y?

Answer 1

Expanding out the left side gives us:\[\large\rm x^2+2xyi-y^2\quad=\quad 15+8i\]

Answer 2

So we have:\[\large\rm \color{orangered}{x^2-y^2}+\color{royalblue}{2xyi}\quad=\quad \color{orangered}{15}+\color{royalblue}{8i}\]The real parts have to match up
while the imaginary parts match up as well, ya?

Answer 3

\[\large\rm x^2-y^2=15\]Applying difference of squares,\[\large\rm (x-y)(x+y)=15\]And we have our other equation as well,\[\large\rm 2xy=8\]

Answer 4

So what can we do with that information... hmmm

Answer 5

what happened to the i again?

Answer 6

Matching up the imaginary parts gives us this,\[\large\rm 2xyi=8i\qquad\implies\qquad 2xy=8\]

Answer 7

u can just equal them like that?

Answer 8

Yes :)
If you have an equivalence like this:\[\large\rm \color{orangered}{x^2-y^2}+\color{royalblue}{2xyi}\quad=\quad \color{orangered}{15}+\color{royalblue}{8i}\]The real on the left must be equal to the real on the right.\[\large\rm \color{orangered}{x^2-y^2=15}\]And the imaginary on the left must be equal to the imaginary on the right.\[\large\rm \color{royalblue}{2xyi=8i\qquad\to\qquad 2xy=8}\]

Answer 9

But I'm getting stuck after that >.<
Hmm thinking...

Answer 10

what if we don't remove any i's, and just do \[(x-y)=\frac{ 15+8i }{ x+y }-2xyi\]

Answer 11

@dan815 @Kainui @Michele_Laino @phi
Hmm :d

Answer 12

@amistre64

Answer 13

zeps idea is more sound to me

Answer 14

x-y = 15/(x+y)

and y=4/x

Answer 15

does x-y have a single solution?

Answer 16

it does not say

Answer 17

the hint that x > 0 seems to imply

(x-y) = 15/(x+4/x)

Answer 18

well, x=4 and y=1 seems to suffice for the constraints, giving x-y = 3

Answer 19

oh yeah that does work. still a bit confused on the work though

Answer 20

zep did most of the process ... if 2 things are equal then their parts must be equal

Answer 21

we could solve the quartic, but I'm guessing there is a tricky way to do this quickly

Answer 22

(x^2-y^2) + 2xy i
===========
15 + 8i

Answer 23

\[ x^2 -y^2 = 15 \\ y= \frac{4}{x} \\ x^2 - \frac{16}{x^2} -15 =0 \]
\[ x^4 -15 x^2 -16= 0 \\ u^2 -15u -16=0 \]
(u-16)(u+1)= 0
u=16
x^2 =16
x=4
y=1
x-y=3

Answer 24

is it possible to solve for the term (x+y) in this problem?

Answer 25

5 would be one solution

Answer 26

but say I had an approach of

(x+y)(x-y)=15+8i-2xyi

xy=4

(x+y)(x-y)=15

could i get x+y just knowing what i have so far?

Answer 27

I only know the way I did it. If we know x and y are real i.e. not imaginary or complex, and x>0, then we would find x=4 , y=1 and x+y= 5
if there is a nicer way to do it, I would like to see it

Answer 28

alright, my teacher will go over this tomorrow so i'll see which method he decides to use. thank you both

Answer 29

if you get any insight, please post it here!