How do I maximize $|t-e^z|$, for $z\in D$, the unit disk? I guess this question doesn't have a closed form solution for all $t\in \Bbb C$, but I know one for $t=1$ provided by Daniel Fischer in a question I asked.
$$\begin{align}
\left\lvert e^w-1\right\rvert &= \left\lvert \sum_{m=1}^\infty \frac{w^m}{m!}\right\rvert\\
&\leqslant \sum_{m=1}^\infty \frac{\lvert w\rvert^m}{m!}\\
&= e^{\lvert w\rvert}-1,
\end{align}$$
with equality for $w \geqslant 0$.
So this $|1-e^z|$ maximized by $z=1$. I tried doing the same for other values of $t$, but without success. Here is an attempt for $t=8$
$$\begin{align}
\left\lvert e^w-8\right\rvert &= \left\lvert \sum_{m=1}^\infty \frac{w^m}{m!} -7\right\rvert\\
&\leqslant \sum_{m=1}^\infty \frac{\lvert w\rvert^m}{m!} + 7\\
&= e^{\lvert w\rvert}+6,
\end{align}$$
But I don't get equality for $w \geqslant 0$. Is there a way to maximize $|t-e^z|$ for $z\in D$, the unit disk, for other values of $t$ besides $0,1$?
 A: Suppose $t=x_0+iy_0$. Then write down the boundary of the image of the unit disk under the map $t-e^z$ in the parametric form for $\varphi \in [0,2\pi)$:
$$
x(\varphi)=x_0-e^{\cos\varphi}\cos{\sin\varphi},
$$
$$
y(\varphi)=y_0-e^{\cos\varphi}\sin{\sin\varphi}.
$$
By the maximum modulus principle the maximum modulus value is obtained somewhere on the boundary. Moreover, tangent line to the boundary at the point of the maximum modulus has to be orthogonal to its radius-vector.
Therefore, we are looking for a (one out of an even number of) solution to the "equation"
$$
(x(\varphi),y(\varphi))\perp\frac{d}{d\varphi}(x(\varphi),y(\varphi)),
$$
which, after substitutions of $x$, $y$ and simplifications is equivalent to
$$
x_0e^{\cos\varphi}(\sin\varphi\cos{\sin\varphi}+\cos\varphi\sin{\sin\varphi})+y_0e^{\cos\varphi}(\sin\varphi\sin{\sin\varphi}-\cos\varphi\cos{\sin\varphi})-e^{2\cos\varphi}\sin{\varphi}=0
$$
The obvious family of cases when this equation has simple solutions is when $y_0=0$. Then $\varphi=0,\pi$ satisfies the equation (one of them is the one we seek). I'd need more time to think whether there are more.
Of course, one has to choose out of all solutions the one with the maximal absolute value, since not every stationary point of the distance function correspond to the maximum. But at least this, I believe, is the easiest way to solve this problem numerically.
A: This is not a full answer, I just wanted to show the plot of the curve $\gamma: \phi \mapsto \exp(\exp(i\phi))$. The solution to the question for a given $t \in {\bf C}$ is given by a point(s) on the curve that are farthest from $t$. There are few observations one can make immediately.


*

*when $t \in {\bf R}$ there are four candidate points, $\phi = 0$, $\phi = \pi$ and two conjugate points, such that the normal to the curve at those point intersects $t$. These are candidate solutions. It seems that when $t < 1$, the solution is $\phi = 0$, when $t > 2$ the solution is $\phi = \pi$, while for $ 1 \leq t \leq 2$ (especially at $t=1.5$) there might be a non-trivial solution $\phi \neq 0, \pi$.



