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Homework Help: Solving complex equations?

  1. Sep 14, 2014 #1
    I have a few complex equations that I am having trouble solving for homework.

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

    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

    2. Relevant 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.

    3. 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!
  2. jcsd
  3. Sep 14, 2014 #2
    That's not right. Set z = 2ixy, then you have z = -z. For what value of z is this true?

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

    Can an absolute value have an imaginary part? What does this tell you about y?
  4. Sep 14, 2014 #3

    Ray Vickson

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    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.
  5. Sep 14, 2014 #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.
  6. Sep 14, 2014 #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.
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