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

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.
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.

• There are not 2 variables. There is one variable, $z$, and one parameter, $\alpha$. The problem is under the form $\left | z^2 + z\times B \right | \leq 1$, where $B$ is a constant. Oct 10 '14 at 12:08
• @Martigan Oh yeah , I'm sorry.What I meant is answer should not be in terms of parameter but rather a number . Oct 10 '14 at 12:13
• It is not possible to have a numerical answer with an undefined $\alpha$ parameter. Otherwise, just take $\alpha=\dfrac{\pi}{2}$! It will simplify everything... (just joking!) Oct 10 '14 at 12:18
• @Martigan Why isn't it? Hypothetically one could calculate all the infinite combinations of z and alpha that satisfy the inequality and find the maximum value of abs. of z. Also I'm fairly sure the question isn't wrong. Oct 10 '14 at 12:24
• Then what your are saying is that $\alpha$ is not a fixed parameter... sorry, that's what I understood from your description. Oct 10 '14 at 12:27

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.

• Okay so alpha is real and alpha ranges over all possible values. Then how to approach the problem? Oct 10 '14 at 12:53
• Thanks for the answer! But I didn't understand the following sentence -"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 your conjencture is indeed correct . Also , I didn't really understand why z has to be real. Also , for solving using calculus , do i have to set the derivative to zero? But by doing that I'm getting z=-cos alpha which is wrong. Oct 10 '14 at 18:53
• No surpise. It was a vague description of intuitive woolgathering in my brain. Let me try again: We've got $(z+B)^2 - B^2$. Think of just real $z$ and $B$ for now. We want to make $z$ big, while keeping this expression small. If you had to choose between $z$ having the SAME sign as $B$ and the opposite sign, you'd choose "opposite" so that the first term would be smaller...allowing you to increase $z$. Now the only question is "do you want B big or small?" If $B$ is zero, then the max $z$ will be $\pm 1$; if $B = 1$, I've already shown $z=-1.5$ words. So big $B$ seems better. Re:calculus: yes. Oct 10 '14 at 19:11
• To expand: be sure you're maximizing $|z|$ under the constraint that $|(z+B)^2 - B^2| \le 1$. That's constrained optimization, and a job for Lagrange multipliers, unless you can do it in some obvious and direct way. Oct 10 '14 at 19:14

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 $z=\sqrt{2}+1$.

• Thanks for the answer! But I've a question: 1. In the first step of triangle inequality , shouldn't the absolute value of LHS be less than what you've written ? (as the 'magnitude' of difference of two sides is less than the third side) Oct 11 '14 at 8:35
• @ A Googler: I have used the fact $|a|-|b|\leq |a+b|$ then because $0\leq |\cos(\alpha)|\leq 1$ the result follows. Oct 11 '14 at 9:46
• Oh yes I got it , fantastic answer! Oct 11 '14 at 11:18