Infimum of complex expression on the unit disc in the complex plane Suppose $|\lambda|<1, Re\lambda < 0, Im\lambda > 0$. What is the largest lower bound independent of $\lambda$ on the expression 
$$\frac{1+|\lambda|^2}{2}+Re\lambda - \frac{2\pi}{n+2}Im\lambda$$ where $n$ is a positive integer? If you like, stipulate that $n$ be as large as you need for the problem to make sense, though make it as small of a lower bound on $n$ as you can, if that makes sense. 
 A: Let us give values to the real and imaginary parts of $\lambda $ to make that expression as small as possible while following your constraints.
The expression is
$$\frac{1+|\lambda|^2}{2}+Re\lambda - \frac{2\pi}{n+2}Im\lambda$$
And we want to minimize all the terms.
$\frac{1+|\lambda|^2}{2}$ is as small as $|\lambda|$ is so make it 0, which is as small as it can get.
$Re\lambda$ cannot go below $-1$ otherwise $|\lambda| > 1$ so make it $-1$.
$ -\frac{2\pi}{n+2}Im\lambda$ gets smaller when $Im\lambda$ gets bigger so let us say $Im\lambda = 1$ since it can't go any bigger: $|\lambda| < 1$. Then the expression becomes 
$$\frac{1+|\lambda|^2}{2}+Re\lambda - \frac{2\pi}{n+2}Im\lambda = \frac{1}{2} - 1 - \frac{2\pi}{n+2} = -(\frac{2\pi}{n+2} + \frac {1}{2})$$ which is the best bound I can think of.
A: I tried to find a lower bound by solving:
$$
\begin{align}
\min_{a,b} \quad & \frac{1 + a^2 + b^2}{2} + a - \frac{2\pi}{n+2}b \\
\textrm{s.t.} \quad & a^2 + b^2 \leq 1 \\
& a \leq 0 \\
& b \geq 0 \\
\end{align}
$$
Note that by taking the closure if the feasible set, I can write $\min$ instead of $\inf$. This problem can be solved for fixed $n$ by any convex optimization solver. I used ipopt. Here is a set of optimal values:
$$
\begin{align}
n & & \textrm{value} \\
1 & & -1.32 \\
2 & & -0.862 \\
3 & & -0.606\\
4 & & -0.448 \\
5 & & -0.344 \\
10 & & -0.129 \\
100 & & -0.00160 \\
1002 & & -0.0000197 \\
\end{align}
$$
In the optimum for $n=1$, $a \approx -0.744$, $b\approx0.668$. As $n\to\infty$, $a \to 1$ and $b \to 0$.
A: This is equivalent to the bivariate real optimization problem:
\begin{array}{ll}
\text{minimize} & \frac{1+x^2+y^2}{2}+x-\frac{2\pi}{n+2} y \\
\text{subject to} & x^2 + y^2 \leq 1
\end{array}
To solve, we build the Lagrangian
$$L(x,y,z) = \frac{1+x^2+y^2}{2}+x-\frac{2\pi}{n+2} y - z(1 - x^2 - y^2)$$
where $z \geq 0$ is the Lagrange multiplier. The optimality conditions are
$$x + 1 + 2zx = 0 \quad y - \frac{2\pi}{n+2} + 2zy = 0 \quad x^2+y^2\leq 1$$
Rearranging,
$$x = \frac{-1}{1+2z} \quad y = \frac{2\pi}{(1+2z)(n+2)}$$
Note that if we were to guess that $z=0$---its smallest legal value---then we'd see that $x=1$ and $x^2+y^2>1$. Therefore it must be the case that $z>0$, and complementary slackness tells us this implies $x^2+y^2=1$. So we solve:
$$x^2 + y^2 = \frac{1}{(1+2z)^2} \left( 1 + \frac{4\pi^2}{(n+2)^2}\right) = 1$$
which leads us to
$$\frac{1}{1+2z} = (1+(4\pi^2)/(n+2)^2)^{-1/2}$$
Substitution yields
$$x = \frac{-1}{(1+(4\pi^2)/(n+2)^2)^{1/2}}, \quad y = \frac{2\pi}{(n+2)(1+(4\pi^2)/(n+2)^2)^{1/2}}$$
Then bring this back into the expression: Since $x^2+y^2=1$, that first term is equal to 1, and we have
$$1 - \frac{1+(4\pi^2)/(n+2)^2}{(1+(4\pi^2)/(n+2)^2)^{1/2}} = 1 - (1+(4\pi^2)/(n+2)^2)^{1/2}$$
