Solving complex number equations for x AND y

In summary, the two equations have the same solutions if y=0, x=1+or-sqr(2), and the real and imaginary parts of x and y are equal.
  • #1
Wardlaw
27
0

Homework Statement



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

Homework Equations





The Attempt at a Solution


 
Physics news on Phys.org
  • #2
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
jegues said:
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?


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
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.

I'm not too sure about this so maybe have someone reapprove my thoughts.

Hope this helps,
 
  • #5
jegues said:
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.

I'm not too sure about this so maybe have someone reapprove my thoughts.

Hope this helps,

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
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
jegues said:
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


I get that, but what happens to the x in the denominator after you equate the real and imaginary parts?
 
  • #8
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
Erm...could you inform me as to where you got the quadratic expression from?
 
  • #10
Here's a different method:

[tex]x+iy=\frac{1}{x-iy}+2[/tex]

multiplying through by [itex]x-iy[/itex] in order to get rid of the denominator

[tex]x^2+y^2=1+2x-i2y[/tex]

Now, remember the rule that [itex]A+Bi\equiv C+Di[/itex] if and only if [itex]A=B[/itex] and [itex]C=D[/itex].
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:
[tex]x^2+y^2=1+2x[/tex] since x and y are real (only the i is imaginary)

the imaginary parts:

[tex]0=-i2y[/tex]
Therefore
[tex]y=0[/tex]

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 [itex]x^2+y^2=1+2x[/itex] 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 [itex]x-iy[/itex] in the denominator, and this cannot be zero, so [itex]x\neq y\neq 0[/itex]. That is, just make sure you don't include the solution x=0 if it is a solution.
 
  • #11
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
jegues said:
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)

It's a common mistake made by those that have only learned 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 learned my lesson.
But still, you should've realized what the OP was trying to show by the way the brackets were placed :smile:
 
  • #13
jegues said:
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)

The solutions to
[tex]x+iy=\frac{1}{x-iy}+2[/tex]
and
[tex]x+iy=\frac{1}{x}-iy+2[/tex]

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
jegues said:
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.

I'm not too sure about this so maybe have someone reapprove my thoughts.

Hope this helps,


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 [tex]\frac{x^{2}-1}{x}=2 \Rightarrow x^{2}-1=2x \Rightarrow x^{2}-2x-1=0[/tex]

There is no denominator to speak of! Then just solve for x.
 
  • #15
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
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.

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

But still, you should've realized what the OP was trying to show by the way the brackets were placed

This is mathematics, I assume nothing :P
 

Related to Solving complex number equations for x AND y

What are complex numbers?

Complex numbers are numbers that have both a real part and an imaginary part. They are represented in the form a + bi, where a and b are real numbers and i is the imaginary unit, equal to the square root of -1.

How do I solve a complex number equation for x and y?

To solve a complex number equation for x and y, you can use a variety of methods such as substitution, elimination, or graphing. However, the most common and efficient method is to use the quadratic formula.

Can complex number equations have more than one solution for x and y?

Yes, complex number equations can have multiple solutions for x and y. This is because there are infinite solutions to the square root of -1, which is represented by i in complex numbers. Therefore, when solving these equations, you may end up with multiple values for x and y.

What is the difference between real and imaginary solutions?

Real solutions are solutions that are represented by real numbers, while imaginary solutions are represented by the imaginary unit i, which is equal to the square root of -1. In complex number equations, both real and imaginary solutions may be present.

How do I check if my solution is correct?

To check if your solution is correct, you can plug the values of x and y back into the original equation and see if it satisfies the equation. If it does, then your solution is correct. Additionally, you can also use a graphing calculator to visually confirm your solution.

Similar threads

  • Calculus and Beyond Homework Help
Replies
6
Views
1K
  • Calculus and Beyond Homework Help
Replies
17
Views
1K
  • Calculus and Beyond Homework Help
Replies
27
Views
802
  • Calculus and Beyond Homework Help
Replies
9
Views
1K
  • Calculus and Beyond Homework Help
Replies
13
Views
1K
  • Calculus and Beyond Homework Help
2
Replies
44
Views
4K
  • Calculus and Beyond Homework Help
Replies
8
Views
2K
  • Calculus and Beyond Homework Help
Replies
19
Views
1K
  • Calculus and Beyond Homework Help
Replies
28
Views
1K
  • Calculus and Beyond Homework Help
Replies
14
Views
1K
Back
Top