06/30/2022, 12:05 AM
Ah unfortunately that's not possible in the general sense, it's very restrictive.
Assume that:
\[
f(z+c) - c = \sum_{j=1}^\infty a_j z^j\\
\]
And
\[
f(z+b) - b = \sum_{j=1}^\infty a_j z^j\\
\]
Then:
\[
f(z) = f(z+b-c) -b+c\\
\]
Then if \(\mu = b-c\) we're restricted to functions such that:
\[
f(z+\mu) = f(z) + \mu\\
\]
Which implies that \(f(z) - z\) is \(\mu\)-periodic.
Assume that:
\[
f(z+c) - c = \sum_{j=1}^\infty a_j z^j\\
\]
And
\[
f(z+b) - b = \sum_{j=1}^\infty a_j z^j\\
\]
Then:
\[
f(z) = f(z+b-c) -b+c\\
\]
Then if \(\mu = b-c\) we're restricted to functions such that:
\[
f(z+\mu) = f(z) + \mu\\
\]
Which implies that \(f(z) - z\) is \(\mu\)-periodic.
