Finding maximum value of absolute value of a complex number given a condition.

Question

On a recent test, I could not solve the following problem:
If $$\left | z^2 + 2zcos\alpha \right | \leq 1 $$ then find the maximum value of absolute value of z. Alpha is not a fixed parameter. Alpha is real and alpha ranges over all possible values. I've never really solved any problem close to this.
Answer should not be in terms of parameter alpha but rather a number, so how do i go about this?
I tried substituting $z= a+ib$ and then finding the abs. value by definition but then it looks like a quartic complex equation under root sign. Please help.

Answer 1

Using the triangle inequality $$|z^2|-|2z\cos(\alpha)|\leq |z^2+2z\cos(\alpha)|\leq 1$$ but $$|z^2|-|2z|\leq |z^2|-|2z\cos(\alpha)|$$ therefore $$|z^2|-|2z|=|z|^2-2|z|\leq 1\Rightarrow (|z|-1)^2\leq2\Rightarrow ||z|-1|\leq\sqrt{2}$$ In other words $$0 \leq |z|\leq \sqrt{2}+1$$ Notice that this inequality is sharp because it is attained for $\alpha=\pi$ and $z=\sqrt{2}+1$.

Answer 2

First, as @Martigan notes, the answer does depend on $\alpha$. For $\alpha = \pi/2$, the answer is $|z| = 1$, while for $\alpha = 0$, the number $z = -1.5$ satisfies the inequality, so the maximum possible value of $|z|$ is at least $1.5$.

If the question is "what's the max absolute value of $|z|$ as $\alpha$ ranges over all possible values?", then setting $\cos(\alpha)$ to $-1$ is probably a good way to find the best answer (although I haven't proved that!), at least in the case where $\alpha$ is restricted to be real.

But whether you're supposed to solve the problem for a fixed $\alpha$ or optimize over all $\alpha$, the problem should say whether $\alpha$ is assumed real or not. If not, you can pick $z = -2\cos(\alpha)$, which gives an expression whose norm is zero, hence certainly less than 1. Since the norm of $\sin \alpha$ is unbounded (for $\alpha \in \mathbb C$), the answer to the "optimize over $\alpha$" version is evidently "infinity".

My conclusion: this is a really badly stated problem.

With the additional hypothesis that $\alpha$ is a fixed real number, I'd write $$ z^2 + 2z \cos \alpha = (z + \cos\alpha)^2 - \cos^2\alpha $$ and play with that. I'd probably to to make a case for why the $z$ that produces the largest-modulus value for this expression has to be real, or better, why it can be rotated to a real number that produces an equally large modulus, by multiplying by $e^{i \theta}$ for some $\theta$. At that point, I'd have a single-variable calculus problem, and I'd solve it.

I can't guarantee that this works, but you asked for help in how to go about this, and I'm telling you what I'd do first. Offhand, it seems like a tough problem for an exam. Then again, I haven't thought much about complex vars since about 1984, so maybe it's easier than I'm making it.

If "$\alpha$ ranges over all possible values" means that I'm to optimize over $\alpha$ as well, I could take the solution to the part I just described (if it's correct!) and optimize over $\alpha$. Or I could go back to my original observation about $-1.5$ and say that $$ (z + \cos\alpha)^2 - \cos^2\alpha $$ seems as if it'd be less than 1 for the largest possible $z$ when the first term is larger than 1 and the second cancels it out. And the most cancellation comes when $\cos^2 \alpha = \pm 1$. Picking $\cos \alpha = -1$, I'd guess that $z = 1 + \sqrt{2}$ provides the optimum.

At this point, having played with the problem in my head, the real work -- proving the conjectured answer correct -- begins.