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Thread: complex = real

  1. #1
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    complex = real

    When I know that a solution of equation is the same in the real numbers and the complex numbers so they are the same?
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  2. #2
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    Re: complex = real

    This makes no sense as written. You are supposed to have a 5 Bagrut in English. Use it.
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  3. #3
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    Re: complex = real

    Quote Originally Posted by romsek View Post
    This makes no sense as written. You are supposed to have a 5 Bagrut in English. Use it.
    In this question the complex and the real is the same because some of the solutions are canceled.
    (1)So the answer if I understand correctly is when the solution is cancelled. But Now I need a guideline that tell me if I am rights.
    The question:
    sqrt (a^2 + b^2) = a + b

    (2)There is another solutions when they are the same when there no cancelling...?
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  4. #4
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    Re: complex = real

    Quote Originally Posted by policer View Post
    In this question the complex and the real is the same because some of the solutions are canceled.
    (1)So the answer if I understand correctly is when the solution is cancelled. But Now I need a guideline that tell me if I am rights.
    The question:
    sqrt (a^2 + b^2) = a + b

    (2)There is another solutions when they are the same when there no cancelling...?
    What do you mean by "cancelled"? The solution provided by romsek in the original post does not involve any "cancelling". He squared both sides. The problem with squaring both sides (when both sides are real) is that you introduce the possibility that one of the sides was negative while the other positive.

    On the other hand, the Complex Square Root finds the first root of unity. So, you still need either $a=0$ or $b=0$. But, you can have $a=x+iy$ with $x> 0$ or $x=0,y\ge 0$ as a valid solution as well. (When you square it, the first root of unity will be the positive real component while the second root of unity will be the negative real component).
    Last edited by SlipEternal; Sep 7th 2018 at 06:58 AM.
    Thanks from policer and topsquark
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  5. #5
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    Re: complex = real

    Quote Originally Posted by SlipEternal View Post
    What do you mean by "cancelled"? The solution provided by romsek in the original post does not involve any "cancelling". He squared both sides. The problem with squaring both sides (when both sides are real) is that you introduce the possibility that one of the sides was negative while the other positive. This is similar to the issue with complex numbers (where squaring may provide more solutions than actually exist). Squaring both sides over the complex numbers is still possible, and any result from that must still hold for the "unsquared" solution. So, squaring both sides gives a necessary condition for a solution, but it does not give a sufficient solution. So, let's look for additional solutions over the complex numbers. We know that either $a$ or $b$ must be zero (that is still a necessary condition for a solution). Now, we want to check if $b=0$, what possible complex values are available for $a$:

    $$\sqrt{(x+iy)^2}=x+iy$$

    This is true when $x\ge 0$.
    Thanks, Thanks...!!!
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  6. #6
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    Re: complex = real

    Quote Originally Posted by policer View Post
    Thanks, Thanks...!!!
    I updated my answer. If $x>0$, then $y$ can be anything. But if $x=0$, then you need $y\ge 0$.
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