I have a well-behaved complex-valued function $f(x)$ (well-behaved in the sense that it can be represented by a Taylor series). Through the addition of a constant, I would like to be able to minimise the modulus of the function over an interval $(x_{1},x_{2})$ in the sense that I want to calculate:
$\min_{C}\left({\max_{x\in(x_{1},x_{2})}\left|C+f(x)\right|}\right)$
The interval itself is small in the sense that $f(x)$ is slowly varying across the interval, but not necessarily so small that I can simply ditch all quadratically small terms in a Taylor expansion about a point within the interval.
Thus, in the first instance one might consider choosing $C=-f\left(\frac{x_{1}+x_{2}}{2}\right)=C_0$, which, to linear order terms in the Taylor expansion of $f(x)$ about $\frac{x_{1}+x_{2}}{2}$, would be the optimal solution (I believe).
However, I would like to know if there is a better solution if the quadratic (or higher order) terms are significant. My attempt so far has been to work in terms of the Taylor expansion of $f(x)$ about $\frac{x_{1}+x_{2}}{2}$ to quadratic order, and I've managed to get out a solution that's rather clunky in the sense that it depends on the quadrant of the complex plane occupied by the ratio of the first two terms in the Taylor expansion.
I would like to know if there's a more elegant way to solve this, perhaps by considering a perturbation about the optimal solution up to linear order, $C=-f\left(\frac{x_{1}+x_{2}}{2}\right)$. Thanks in advance for any help.
EDIT: Still uncertain how to analytically obtain a solution for complex-functions
When $f(x)$ is complex-valued
I believe $(x_1,x_2)$ should map to a curve in the complex plane. Determine the circum-circle for this mapping; i.e. the smallest circle that contains the mapping. Then the minimizing $C$ is the negative of the centre of this circle, and $\min_C \max_{x\in(x_{1},x_{2})} |C + f(x)|$ is the radius of the circle. The logic is that $\max _x|f(x)|$ is the radius of the circumcircle over $f(x)$ and centred at origin. So to minimize this, shift the mapping (by $C$) to be around the origin.
When $f(x)$ is real-valued
$-C$ marks the new '$0$' value for the expression $|C + f(x)|$. The maximum of this expression then is the largest distance from this zero. To minimize this distance, then put the new zero baseline halfway between the min and max of the function, $$ C = -\frac{f^{\max} + f^{\min}}{2} \rightarrow \min_C \max_{x\in(x_{1},x_{2})} |C + f(x)| = \frac{f^{\max} - f^{\min}}{2} \\ f^{\min} \equiv \min_{x\in(x_{1},x_{2})} f(x), \; f^{\max} \equiv \max_{x\in(x_{1},x_{2})} f(x) $$
The following is the algebraic derivation, $$ C + f^{\min} \leq C + f(x) \leq C + f^{\max} \\ |C + f(x)| \leq \max \left( |C + f^{\min}|, |C + f^{\max}|\right) \\ \implies \max_{x\in(x_{1},x_{2})} |C + f(x)| = \max \left( |C + f^{\min}|, |C + f^{\max}|\right) \\ $$ $$ C \geq - f^{\min} \rightarrow \max_{x\in(x_{1},x_{2})} |C + f(x)| = |C + f^{\max}| \geq f^{\max} - f^{\min}\\ C \leq - f^{\max} \rightarrow \max_{x\in(x_{1},x_{2})} |C + f(x)| = |C + f^{\min}| \geq f^{\max} - f^{\min} \\ -f^{\max}\leq C \leq - f^{\min} \rightarrow \max_{x\in(x_{1},x_{2})} |C + f(x)| = \max \left( - C - f^{\min}, C + f^{\max}\right) \leq f^{\max} - f^{\min} \\ \therefore \min_C \max_{x\in(x_{1},x_{2})} |C + f(x)| \implies -f^{\max}\leq C \leq - f^{\min}\\ $$
$$ a(C) \equiv - C - f^{\min},\; b(C) \equiv C + f^{\max} \\ \rightarrow \min_C \max_{x\in(x_{1},x_{2})} |C + f(x)| = \min_C \max(a(C),b(C)) \\ \text{where} \; a + b = f^{\max} - f^{\min}, \; 0 \leq a,b \leq f^{\max} - f^{\min} \\ \therefore C = -\frac{f^{\max} + f^{\min}}{2} \rightarrow \min_C \max_{x\in(x_{1},x_{2})} |C + f(x)| = \frac{f^{\max} - f^{\min}}{2} $$