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

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In summary: I'm not sure how to do the same for the y = 0.In summary, the student is having trouble solving equations involving the real and imaginary parts. He is able to solve equations involving the real and imaginary parts for the real parts, but is having trouble solving equations involving the real and imaginary parts for the imaginary parts. He is able to solve equations involving the real and imaginary parts for the real parts, but is having trouble solving equations involving the real and imaginary parts for the imaginary parts. He is able to solve equations involving the real and imaginary parts for the real parts, but is having trouble solving equations involving the real and imaginary parts for the imaginary parts.
  • #1
159753x
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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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  • #2
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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  • #3
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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  • #4
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.
 
  • #5
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.
 

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

What is the general process for solving complex equations?

The general process for solving complex equations involves isolating the variable on one side of the equation and performing inverse operations until the variable is left alone. This includes using properties of equality and solving for a single variable at a time.

How can I simplify complex equations before solving them?

To simplify complex equations, you can use the distributive property, combine like terms, and remove parentheses. It is also helpful to convert all terms to the same base or eliminate fractions by multiplying by the least common denominator.

What are some common mistakes to avoid when solving complex equations?

Some common mistakes to avoid when solving complex equations include not distributing negative signs correctly, not following the correct order of operations, and making errors when simplifying or combining like terms.

How do I know when a complex equation has no solution?

If an equation results in a contradiction, such as 5 = 6 or 0 = 1, then it has no solution. Additionally, if the variables cancel out on both sides of the equation, it means there are no solutions that satisfy the equation.

What are some tips for solving complex equations with multiple variables?

When solving equations with multiple variables, it is important to isolate one variable at a time and use substitution to solve for the remaining variables. It can also be helpful to use elimination or substitution methods to eliminate one variable and solve for the others.

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