• Support PF! Buy your school textbooks, materials and every day products Here!

Determing whether this function is onto and one to one

  • Thread starter bonfire09
  • Start date
  • #1
248
0

Homework Statement


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].


Homework Equations





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?.
 

Answers and Replies

  • #2
34,231
10,272
Does that imply its not one to one and onto in the complex numbers?
One of those is implied, but the other is not.
You can use real numbers to check this.

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?.
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)
 
  • #3
248
0
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:
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
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.
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.
 
  • #5
248
0
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.
 
  • #6
Dick
Science Advisor
Homework Helper
26,258
618
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.
Sure, but x^2 IS onto in the complex numbers. Not the reals. That's what I meant.
 
  • #7
34,231
10,272
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.
 
  • #8
248
0
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.
 
  • #9
34,231
10,272
Can you solve z=(x+1)^2 for x (or at least one option for x)? Is this possible for all complex z?
 
  • #10
248
0
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?
 
  • #11
34,231
10,272
For every z, you found (at least) one x such that f(x)=z. That is sufficient.
 

Related Threads on Determing whether this function is onto and one to one

Replies
1
Views
966
  • Last Post
Replies
3
Views
978
  • Last Post
Replies
0
Views
1K
Replies
5
Views
2K
  • Last Post
Replies
5
Views
764
Replies
14
Views
2K
  • Last Post
Replies
16
Views
11K
Replies
4
Views
845
Replies
2
Views
6K
Top