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. 

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
$$1 - \frac{1+(4\pi^2)/(n+2)^2}{(1+(4\pi^2)/(n+2)^2)^{1/2}} = 1 - (1+(4\pi^2)/(n+2)^2)^{1/2}$$
