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Determing whether this function is onto and one to one

  1. Sep 6, 2013 #1
    1. The problem statement, all variables and given/known data
    Determine whether the function is one to one and whether it is onto. The function is [tex] f:\mathbb{C}\rightarrow\mathbb{C}: f(x)=x^2+2x+1[/tex].


    2. Relevant equations



    3. The attempt at a solution
    Now I know this function in the real numbers is neither one to one and onto. Does that imply its not one to one and onto in the complex numbers? And what if I did have a one to one and onto function in the complex numbers. I can't think of a example but how would I show that?.
     
  2. jcsd
  3. Sep 6, 2013 #2

    mfb

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    One of those is implied, but the other is not.
    You can use real numbers to check this.

    If you need an example: f(x)=x.
    You can show it in the same way as for real functions, you just have to work with complex numbers (so equations like x2=-1 have solutions, for example)
     
  4. Sep 6, 2013 #3
    So I want to showf(x)=x^2 is not one to one in the complex number so then id suppose [itex] x_1=x_2[/itex] then let [itex] x_1,x_2\in\mathbb{C}[/itex] where [itex]x_1=a+bi (and) x_2=c+di[/itex] where [itex]a,b,c,d\in\mathbb{C}[/itex]. So [itex] f(x_1)=(a+bi)^2=a^2+2abi-b^2[/itex] and [itex]f(x_2)=(c+di)^2=c^2+2cdi-d^2[/itex]. But then i don't know what to do from here? Because c and d could be different from a and b or the same.
     
    Last edited: Sep 6, 2013
  5. Sep 6, 2013 #4

    Dick

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    If the function isn't one to one in the reals, it's not one to one in the complex numbers. The reals are basically a subset of the complex numbers. It might be onto. Use polar representation and Euler's formula to show every complex number has a square root.
     
  6. Sep 6, 2013 #5
    Oh OK. I've never really done that before but I'll mess around with it and see. But if you look in the real' numbera f(x)=x^2 is not onto either since x^2 is always greater than or equal to 0.
     
  7. Sep 6, 2013 #6

    Dick

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    Sure, but x^2 IS onto in the complex numbers. Not the reals. That's what I meant.
     
  8. Sep 7, 2013 #7

    mfb

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    Did you recognize the binomial formula in your function? This simplifies the problem.
    To show that it is not one to one, you can pick some function value (like 4) and show that there are two x such that f(x)=4.
     
  9. Sep 8, 2013 #8
    I got the 1 to 1 part figured out for f(x)=(x+1)^2. I just gave a counterexample in the reals.' Now I'm stuck trying to prove it is onto.. I'm not sure how to convert this equation into polar form.
     
  10. Sep 8, 2013 #9

    mfb

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    Can you solve z=(x+1)^2 for x (or at least one option for x)? Is this possible for all complex z?
     
  11. Sep 8, 2013 #10
    Yes just by looking at that form there is always going to be atleast one root. So I should solve that for x then and which would give me [itex] x=\sqrt{z}-1[/itex]? Then plug it back in and show it does give me back the z?
     
  12. Sep 8, 2013 #11

    mfb

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    For every z, you found (at least) one x such that f(x)=z. That is sufficient.
     
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