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Let $g$ be a complex number, where $a$, $b$, $c$, $d$ are real numbers, and $i = \sqrt{-1}$.

$g = \frac{{(a + bi)\exp (ci)}}{{{\rm{abs}}(a + bi)}}$

Since the absolute value (i.e. modulus) of a product of complex numbers is the product of the absolute values:

${\rm{abs((a + b}}i{\rm{)(c + d}}i{\rm{)) = abs(a + b}}i{\rm{)abs(c + d}}i{\rm{)}}$

In addition,

$\exp (ix) = \cos (x) + {\rm{ }}i\sin (x)$

${\rm{abs}}(\exp (ix)) = 1$

So from the above,

${\rm{abs(g) = 1}}$

I am searching for a Multiplicative function $f(g)$ such that:

$f(g) \neq 1$

$f(g) = f\left( {\frac{{(a + bi)}}{{{\rm{abs}}(a + bi)}}} \right)f\left( {\exp (ci)} \right) \ne 1$


$f\left( {\frac{{(a + bi)}}{{{\rm{abs}}(a + bi)}}} \right) = 1$


$f\left( {\exp (ci)} \right) = 1$


$f\left( {\frac{{(a + bi)}}{{{\rm{abs}}(a + bi)}}} \right) \ne f\left( {\exp (ci)} \right)$

Does such a function exist? Why or why not? What if $a,b,c$ are not known, and only $g$ is known?

Note that it appears that the complex logarithm (Wikipedia) of $g$ will always have a real part equal to zero, so that $\log(g)$ has only an imaginary part:

$\log (g) = \log \left( {\frac{{a + bi}}{{{\rm{abs(}}a + bi{\rm{)}}}}} \right) + \log (\exp (ci))$

$\log (g) = \log ({\rm{abs(}}a + bi)) + \arg (a + bi)i - \log ({\rm{abs}}(a + bi)) + \log (\exp (ci))$

$\log (g) = \arg (a + bi)i + \log (\exp (ci))$

If there is a logarithm to another base $k$, such that ${\log _k}\left( {\frac{{a + bi}}{{{\rm{abs(}}a + bi{\rm{)}}}}} \right)$ is real and ${\log _k}\left( {\exp (ci)} \right)$ is complex (or vice versa), then it might be possible to separate ${\frac{{a + bi}}{{{\rm{abs(}}a + bi{\rm{)}}}}}$ from ${\exp (ci)}$.

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If you assume $g\neq 1,$ then complex conjugation, or even identity, works. – Andrew Sep 14 '12 at 18:35
I think he actually wants a group homomorphism $f: U(1) \to \mathbb{C}^*$ so that $Im(f) \nsubseteq U(1)$. – N. S. Sep 14 '12 at 18:38
N.S. Yes, N.S., I think that is a very elegant way of saying it, and Andrew's comment is correct as well. Ideally I would like $f\left( {\frac{{(a + bi)}}{{{\rm{abs}}(a + bi)}}} \right) = 1$ or $f\left( {\exp (ci)} \right) = 1$ – Nicholas Kinar Sep 14 '12 at 18:49
Can such a group homomorphism be constructed? Is there a good book that can be used as a reference? – Nicholas Kinar Sep 14 '12 at 19:42
up vote 1 down vote accepted

If you want this to hold all the time (i.e. for any $a,b,c$ excluding $a=b=0$), then your second point has to be relaxed. $\frac{a+bi}{|a+bi|}$ and $e^{ci}$ are both points on the unit circle. So if one of those values maps to $1$, the other can have its constants chosen so that it is the same value, and so is also mapped to one. (e.g, with $a=0,b=1$ we choose $c=\pi/2$). So if the product can never map to $1$, then nothing can map to $1$. At the moment I'm unclear as to why something such as $f(z)=2|z|$ wouldn't suit your purposes, since that seems a bit simple for the complexity of the question.

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Thanks for your response. Ideally, I would like $f\left( {\frac{{(a + bi)}}{{{\rm{abs}}(a + bi)}}} \right) = 1$ or $f\left( {\exp (ci)} \right) = 1$, if possible. – Nicholas Kinar Sep 14 '12 at 18:52
That's the thing: if $f(e^{ic})$=1, I can pick $a,b$ such that $\frac{a+bi}{|a+bi|}=e^{ic}$, so that the whole product will be $1$. If anything ever maps to one, then a product exists that will as well. So if the product can never map to one, then nothing can. – Robert Mastragostino Sep 14 '12 at 18:59
Thanks Robert. Okay, but what if I cannot pick $a,b$, and these are "given"? Is there still a transformation $f(x)$ with the properties that I am searching for? – Nicholas Kinar Sep 14 '12 at 19:09
@NicholasKinar if they're given, you can pick $c$ such that they're equal and you get the same problem. If you can't pick anything, then it's trivial because you're only dealing with a domain of three numbers: just pick $f(\frac{a+bi}{|a+bi|})=m, f(e^{ic})=n$ and define $f(\frac{(a+bi)e^{ic}}{|a+bi|})=mn$. – Robert Mastragostino Sep 14 '12 at 19:43
If I'm given $g = \frac{{(a + bi)\exp (ci)}}{{{\rm{abs}}(a + bi)}}$ without knowing $a,b,c$, can I still find an $f(x)$ such that $f(g) = f\left( {\frac{{(a + bi)}}{{{\rm{abs}}(a + bi)}}} \right)$ or $f(g) = f\left( {\exp (ci)} \right).$ Thanks again, Robert. – Nicholas Kinar Sep 14 '12 at 19:50

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