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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:

$f(g) \neq 1$

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

Ideally,

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

or

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

but

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

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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    Can such a group homomorphism be constructed? Is there a good book that can be used as a reference?2012-09-14

1 Answers 1

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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 Robert. This answer contains all of the information.2012-09-16