Prove that if $f=u+iv$ is regular through out the complex plane and $au+bv-c\geq0$ for some suitable constants $a,b,c$ then $f$ is constant.
2026-03-29 05:34:04.1774762444
Entire function being constant if $au+bv-c\geq0$
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I suppose that you are assuming that $(a,b)\neq(0,0)$; otherwise, the statement is trivially false.
Let $g=(-a+bi)f+c$. Since $f=u+vi$, $g=-au-bv+c+i(-av+bu)$, and so $\operatorname{Re}(g)=-au-bv+c\leqslant0$. Therefore, $e^g$ is bounded (since $|e^g|=e^{\operatorname{Re} g}\leqslant1$, and then Liouville's theorem implies that $e^g$ is constant. Can you take it from here?