Solving Complex Equations: How Can We Separate Real and Imaginary Components?

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Homework Help Overview

The discussion revolves around solving complex equations involving real and imaginary components. The original poster presents three specific equations and expresses difficulty in separating and solving for the real numbers x and y.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the method of separating real and imaginary parts of complex equations. The original poster attempts to apply this method but encounters confusion regarding the implications of their results, particularly in the first equation.
  • Some participants question the original poster's interpretations and suggest reconsidering the steps taken, especially regarding the handling of imaginary components and absolute values.
  • There is a focus on understanding the conditions under which certain equations hold true, particularly in relation to the values of x and y.

Discussion Status

The discussion is ongoing, with some participants providing guidance on the interpretation of results and the implications of the equations. The original poster has made progress on some equations but seeks further clarification on the last equation, indicating a productive exchange of ideas.

Contextual Notes

Participants note the importance of correctly interpreting the results of separating real and imaginary components, as well as the implications of absolute values in complex equations. There is an acknowledgment of potential oversights in the original poster's approach.

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I have a few complex equations that I am having trouble solving for homework.

Homework Statement



Solve for all possible values of the real numbers x and y.

A. (x+iy)2 = (x-iy)2

B. (x + iy + 2 + 3i)/(2x + 2iy - 3) = i + 2

C. Abs[1 - (x + iy)] = x + iy

Homework Equations



The example problem in the book says that we should separately solve the real and complex parts. That is what I try to do.

The Attempt at a Solution



A. Expanding both sides, I simply get x2 - y2 = x2 - y2 for the real parts. I don't know what to do with that information.

For the imaginary parts, I get 2ixy = -2ixy. So I get plus-or-minus y = plus-or-minus x.

The answer in the back is x = 0 for any real y OR y = 0 for any real x. How did they get this?

B. Again, separating out the real and imaginary components:

(x + 2)/(2x - 3) = 2
Solving this, I get 8/3.

For the imaginary part, I get (y + 3)/(2y) = 1. This yields y = 3.

The answer in the back is x = 36/13 and y = 2/13.

C. I don't know how to deal with the absolute value in this one. The answer is y = 0, x = 1/2.I solved many other problems using the separation of real and imaginary components strategy, but these don't seem to work. Some help would be appreciated!
 
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159753x said:
For the imaginary parts, I get 2ixy = -2ixy. So I get plus-or-minus y = plus-or-minus x.

That's not right. Set z = 2ixy, then you have z = -z. For what value of z is this true?

159753x said:
B. Again, separating out the real and imaginary components:

(x + 2)/(2x - 3) = 2

No, think about what you are doing - the first step is to multiply both sides by (2x + 2iy -3)

159753x said:
C. I don't know how to deal with the absolute value in this one.

Can an absolute value have an imaginary part? What does this tell you about y?
 
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159753x said:
I have a few complex equations that I am having trouble solving for homework.

Homework Statement



Solve for all possible values of the real numbers x and y.

A. (x+iy)2 = (x-iy)2

B. (x + iy + 2 + 3i)/(2x + 2iy - 3) = i + 2

C. Abs[1 - (x + iy)] = x + iy


Homework Equations



The example problem in the book says that we should separately solve the real and complex parts. That is what I try to do.

The Attempt at a Solution



A. Expanding both sides, I simply get x2 - y2 = x2 - y2 for the real parts. I don't know what to do with that information.

For the imaginary parts, I get 2ixy = -2ixy. So I get plus-or-minus y = plus-or-minus x.

The answer in the back is x = 0 for any real y OR y = 0 for any real x. How did they get this?

B. Again, separating out the real and imaginary components:

(x + 2)/(2x - 3) = 2
Solving this, I get 8/3.

For the imaginary part, I get (y + 3)/(2y) = 1. This yields y = 3.

The answer in the back is x = 36/13 and y = 2/13.

C. I don't know how to deal with the absolute value in this one. The answer is y = 0, x = 1/2.


I solved many other problems using the separation of real and imaginary components strategy, but these don't seem to work. Some help would be appreciated!

You have made a very basic error in (A): from the equation ##2 i xy = -2i x y## you can conclude that ##xy = 0##, so either ##x=0## or ##y = 0## or both. Think about what these mean.
 
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Thanks to both of you, Mr Anchovy and Ray Vickson.

I was able to solve both A and B.

It is now clear to me that 2ixy = 2ixy (ie. z= -z) only when either x or y (or both) = 0. This is obvious now! :smile:

The same is true for B. That was a silly oversight. I ended up with the correct answers after multiplying both sides of the original equation by the denominator on the left side. The answers are x = 36/13 and y = 2/13.

The last one is a little tricker. I now remember that the absolute value of z is just Sqrt[z z*], so doing that I was able to get x = 1/2 from solving the real part of the equation.

However, after expanding everything and simplifying, all I am left with is 0 = 2xy. Intuitively, it would seem that x or y or both need to be zero to satisfy this expression. I thought this would be the answer, but instead the answer says "x = 1/2, y = 0". I can get the x = 1/2, but the y = 0 doesn't make sense to me right away.

Could you please help me with this? I am retrying the problem and will let you know if another idea comes up.
 
What is ## 2 \times 0.5 \times 0 ##?

Alternatively, as I hinted before, because Abs is a real-valued function, the imaginary part of its value (i.e. the coefficient of i in x + iy) must be zero.
 

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