Question

Solve the equation ((-i)/(x-iy)) = ((4+7i)/(5-3i)) for x and y, leaving your answers as rational numbers.

I made the denominator of both real so

((-x)/(x^2+y^2)) = (47/34)
and
((y)/(x^2 + y^2)) = (-1/34)

however I don't know where to go from here.

Any help would be appreciated.

Answer 1

Forgot to mention I was comparing the real and imaginary parts for the second bit.

Answer 2

I got -47/34 and -1/34....

Answer 3

I have -x on the numerator, I just didn't simplify that when I typed it.
How would I go about solving for x and y? I seem to go in circles.

Thanks

Answer 4

Ah, Ok we agree then... let me have a look.

Answer 5

It's a non-calculator question.
I probably should've mentioned that.

Answer 6

x^2 + y^2 = 1 ?

Answer 7

Unit circle?

Answer 8

Where did you get that from?

Answer 9

It's a standard equation for a circle with radius 1.

Or parametrized, cos^theta + sin^2 theta = 1.

Answer 10

How did you get to x^2 + y^2 = 1?
I can't get out of having y's. So, I have x^2 + y^2 = -34/y

Answer 11

-34y sorry.

Answer 12

You can regard your complex numbers as living in the complex plane.

I'm not sure what you mean by a derivation.

If I take a point x,y on the unit circle in the first quadrant, pythagorus tells u that x is cos theta and y is sin theta and cos^2 + sin^2 is 1.

U can also get it from de Moivre's theorem.

But it is accepted as standard.

Answer 13

So x is 47/34 and y is -1/34, u already found the answer.

Answer 14

-47/34, sorry.

Answer 15

If
((x)/(x^2+y^2)) = (-47/34)
and
((y)/(x^2 + y^2)) = (-1/34)

Then surely x and y themselves will have different values.

As x = (-47/34)(x^2+y^2) and y = (-1/34)(x^2 + y^2)

Plugging those into WolframAlpha I get x as -47/65 and y as = -1/65

I can't work how to solve the equations for x and y however.

Answer 16

http://www.wolframalpha.com/input/?i=%28%28-i%29%2F%28x-iy%29%29+%3D+%28%284%2B7i%29%2F%285-3i%29%29

Answer 17

U want x and y real, right?

Answer 18

Yes

Answer 19

So yuo scroll down the page to the alternate form (x, y real).

Answer 20

That hasn't solved for x and y, that has just stated that
((ix)/(x^2+y^2)) = (-47i/34) and ((y)/(x^2 + y^2)) = (-1/34).

Answer 21

That is telling you (by comparing real and imaginary parts as u did) that the answer is an alternate form of your equation. The x^2 +y^2 are gone because they equal 1.

Not sure about the i artifact, could be because of the way u have written the question with -i at the beginning.

Answer 22

Notice u get the same result if swap i for -i

Answer 23

Let me try another, simpler tack.

If I write 2x/5 = 3y/5 = 1/5 do u agree that I can cancel the 5's

Answer 24

Treating the compared real and imaginary parts (without the i's) as simultaneous equations and plugging them in gets

http://www.wolframalpha.com/input/?i=x+%3D+%28-47%2F34%29%28x^2%2By^2%29%2C+y+%3D+%28-1%2F34%29%28x^2+%2B+y^2%29

Which is the value for x and y.

Answer 25

Also, yes.
It's multiplying through by 5.

Answer 26

I don't know what else to tell u, I think u have the answer already.

Answer 27

Leave the question here, see if someone else will chime in....

Answer 28

Ok, thanks for the help anyway.

Answer 29

Alright, I figured this out now....

Answer 30

The thing goes awry in your method.

Simply cross multiply and multiply out.

Then compare your real and imaginary parts and u get 2 simultaneous equations in x and y.

Answer 31

I have 2 simultaneous equations at the moment which I can't solve. I can't get out of being in x's and y's.

Are they different doing it that way? Generally, it seems to be better to make the denominator real first.

Answer 32

No, that's what causes all the problems (the x^2 + y^2).

Just do what I said you will see--y =-1/65, I didn't calculate x.

Answer 33

Alright, I'll try it. Thanks.

Answer 34

Ok, I've got it now.
Thank you.

Wish I tried that earlier though.

Answer 35

U and me both...:-)

Still, good workout.

Answer 36

Yep.