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 ${\exp (ci)}$.