# Homework Help: Solving complex number equations for x AND y

1. Dec 24, 2009

### Wardlaw

1. The problem statement, all variables and given/known data

(x+iy)=[1/x-iy]+2

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Dec 24, 2009

### jegues

I'm confused at what your trying to do here? Could you state the question you were given for us? Your attempt at the solution?

3. Dec 24, 2009

### Wardlaw

Okay, sure.
I was asked to equate the real and imaginary parts and solve the the equationi gave for x and y. My solution was to use the complex conjugate and tryto solve that way, however that did not work. Any suggestions?

4. Dec 24, 2009

### jegues

Hmmm... I'm not sure if this right but here's what I did.

I first grouped all the real and imaginary parts together on the LHS of the equation giving me the following

[ (x^2-2x-1) / x ] + 2iy = 0

You can use quadratic formula to solve the roots of the numerator in the real part and as for the imaginary parts you can simply set y = 0.

So, x = 1 +or- sqr(2) and y = 0 should satisfy the equality if I'm not mistaken.

Hope this helps,

5. Dec 24, 2009

### Wardlaw

Okay, at first glance i thought that was the solution. Would it be at all possible for you to show me the working for the quadratic solutions?

6. Dec 24, 2009

### jegues

In the numerator your have a quadratic of the form Ax^2-Bx-C, so to find its roots just simply apply the quadratic formula.

So, x = ( -B +or- sqr[ (B)^2 - 4(A)(C) ] ) / 2A

7. Dec 24, 2009

### Wardlaw

I get that, but what happens to the x in the denominator after you equate the real and imaginary parts?

8. Dec 24, 2009

### jegues

Who cares about the x in the denominator ;)!

As long as its not equal to 0 the result will remain the same. The numerator is zero, so any real number in the world and sit in the denominator we don't care, its still going to go to zero.

0 divided by a number is going to be 0.

9. Dec 24, 2009

### Wardlaw

Erm...could you inform me as to where you got the quadratic expression from?

10. Dec 24, 2009

### Mentallic

Here's a different method:

$$x+iy=\frac{1}{x-iy}+2$$

multiplying through by $x-iy$ in order to get rid of the denominator

$$x^2+y^2=1+2x-i2y$$

Now, remember the rule that $A+Bi\equiv C+Di$ if and only if $A=B$ and $C=D$.
This means that the real parts on the left side of the equation must be equal to the real parts on the right side, and the same goes for the imaginary parts.

So, the real parts:
$$x^2+y^2=1+2x$$ since x and y are real (only the i is imaginary)

the imaginary parts:

$$0=-i2y$$
Therefore
$$y=0$$

So now we know the solution(s) to this equation must be on the line y=0 (which means the solutions are purely real). But we need to look at the first equation $x^2+y^2=1+2x$ and solve this to find what the actual solutions are. Remember, y=0 so you can substitute that into this equation, and when you find your solutions, remember to check the actual question again because a solution might be invalid:

Since, in the original question there is $x-iy$ in the denominator, and this cannot be zero, so $x\neq y\neq 0$. That is, just make sure you don't include the solution x=0 if it is a solution.

11. Dec 24, 2009

### jegues

Mentallic has the right solution, you didn't use brackets properly in your OP so saw it as 1/x - iy, not the proper form,

1/(x-iy)

12. Dec 25, 2009

### Mentallic

It's a common mistake made by those that have only learnt to use the new-age calculators that can display the fraction properly. "Back in my day" - a few years ago, I had to actually throw those brackets in for a fraction, so I learnt my lesson.
But still, you should've realized what the OP was trying to show by the way the brackets were placed

13. Dec 25, 2009

### Mentallic

The solutions to
$$x+iy=\frac{1}{x-iy}+2$$
and
$$x+iy=\frac{1}{x}-iy+2$$

are exactly the same. haha that would've been bad for OP to find that the solutions you've shown are correct, so then he goes ahead and uses that method for every other question of this type.

14. Dec 25, 2009

### yungman

This is absolutely right, there is no other way to put it, group all the real part and immaginary part and solve the real part and immaginary separately. In this case, y=0.

On the real part is $$\frac{x^{2}-1}{x}=2 \Rightarrow x^{2}-1=2x \Rightarrow x^{2}-2x-1=0$$

There is no denominator to speak of!!! Then just solve for x.

15. Dec 25, 2009

### Count Iblis

You can also sokleve this as follows. If we put z = x + i y, then we have upon multiplying both sides by z*:

|z|^2 = 1 +2 z* --->

We conclude from this that z* is real. So, z is real and then you just have to solve the quadratic equation to find x.

16. Dec 26, 2009

### jegues

There's nothing wrong with the method I used, he just didn't display the question properly :P

This is mathematics, I assume nothing :P