Infimum of complex expression on the unit disc in the complex plane

Question

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.

Answer 1

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

Answer 2

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.

Answer 3

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 $b \to 0$.