Construct a complex function with these properties

[chipotleaway]

Homework Statement

Construct a function $f:C \rightarrow C$ such that $f(x+y)=f(x)+f(y) and f(xy)=f(x)f(y)$ (aside from the identity function) Hint: $i^2=-1$ what are the possible values of $f(i)$.

The Attempt at a Solution

All I've been able to do so far is come up with some (hopefully correct) examples e.g.
$f(i^2)=f(-1)=f=((1,0)*(-1,0))=f(-1)f(1)$

$f(a+bi)=f((a,0)+(0,b))=f(a)+f(bi)$

$f(ai)=f((a,0)*(0,1))=f(a,0)f(0,1)=f(a)f(i)$ (a is real)

But I'm not sure how to construct an explicit function out of this (I'm assuming f(x)=0 would be a 'trivial' construction?)

 

[Simon Bridge]

You only need to find a function - so try a few simple non-trivial functions and see how they behave.

i.e. f(x)=x ... what happens?
hint: make x and y general complex numbers so x=a+ib, y=c+id.

you basically need to play around until you notice a pattern, then exploit the pattern.
enjoy.

 

[pasmith]

Homework Statement

Construct a function $f:C \rightarrow C$ such that $f(x+y)=f(x)+f(y) and f(xy)=f(x)f(y)$ (aside from the identity function) Hint: $i^2=-1$ what are the possible values of $f(i)$.

The Attempt at a Solution

All I've been able to do so far is come up with some (hopefully correct) examples e.g.
$f(i^2)=f(-1)=f=((1,0)*(-1,0))=f(-1)f(1)$

I think the point is that $f(i)^2 = f(i^2) = f(-1)$.

The condition $f(xy) = f(x)f(y)$ requires that
$$f(x) = f(1x) = f(1)f(x)$$
so that
$$f(x)(1 - f(1)) = 0$$
for all $x$. From this one can find $f(0)$, and then $f(-1) = f(0) - f(1)$.

 

[BruceW]

another interesting question, is what are all possible functions that satisfy these requirements. But as Simon says, for this problem you just need to find one. I'd also agree with Simon, to just play around for a while longer. For example, you already have f(-1) = f(-1) * f(1) And you can find more of these simple relations, you should eventually be able to guess what the function is.

 

[chipotleaway]

Another relation I got is f(x)=nf(1)f(x) for any natural number n, because f(x)=f(1*x)=f(1)f(x)=f(1*1)f(x)=f(1)f(1)f(x) and you can just keep doing this. Is this correct? Seems like a pretty strong condition

 

[Dick]

Another relation I got is f(x)=nf(1)f(x) for any natural number n, because f(x)=f(1*x)=f(1)f(x)=f(1*1)f(x)=f(1)f(1)f(x) and you can just keep doing this. Is this correct? Seems like a pretty strong condition

You mean f(x)=f(1)^n*f(x) (not nf(1)f(x)). That's not too hard to satisfy if f(1)=1.

 

[pasmith]

Another relation I got is f(x)=nf(1)f(x) for any natural number n, because f(x)=f(1*x)=f(1)f(x)=f(1*1)f(x)=f(1)f(1)f(x) and you can just keep doing this. Is this correct? Seems like a pretty strong condition

That would give you $f(x) = f(1)^n f(x)$.

Did it not occur to you to stop after $f(x) = f(1x) = f(1)f(x)$ and see what that requires of $f(1)$? See my earlier post.

You also have
$$f(0) = f(0\times 1) = f(0)f(1)$$
and
$$f(0) = f(0 \times 2) = 2f(0)f(1)$$
What does
$$f(0)f(1) = 2f(0)f(1)$$
require of $f(0)f(1)$?

 

[chipotleaway]

Haha, brain malfunction there.

Anyway, I got it now (quite certain it didn't take anyone else half as long to do it, don't think they needed this much help either!)

f(1)=1, f(a)=a for any real a because f(a)=f(1+...+1)=f(1)+..+f(1)=1+..+1=a (a times).

For the imaginary part, f(i^2)=f(i)^2=f(-1)=-1 (by the above). So f(i) is i or minus i.

f(a+bi)=f(a)+f(b)f(i)=a+bi if i is positive, but that's the identity function so it must be f(a+bi)=a-bi.

Thanks a lot everyone

 

[BruceW]

very nice! although really, f(a)=a for any rational a (and this solution is not a unique solution, for 'a' real). But I think that all the other solutions are highly pathological anyway.