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.

  • $\begingroup$ 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. $\endgroup$
    – Martigan
    Oct 10 '14 at 12:08
  • $\begingroup$ @Martigan Oh yeah , I'm sorry.What I meant is answer should not be in terms of parameter but rather a number . $\endgroup$
    – A Googler
    Oct 10 '14 at 12:13
  • $\begingroup$ 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!) $\endgroup$
    – Martigan
    Oct 10 '14 at 12:18
  • $\begingroup$ @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. $\endgroup$
    – A Googler
    Oct 10 '14 at 12:24
  • $\begingroup$ Then what your are saying is that $\alpha$ is not a fixed parameter... sorry, that's what I understood from your description. $\endgroup$
    – Martigan
    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.

  • $\begingroup$ Okay so alpha is real and alpha ranges over all possible values. Then how to approach the problem? $\endgroup$
    – A Googler
    Oct 10 '14 at 12:53
  • $\begingroup$ 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. $\endgroup$
    – A Googler
    Oct 10 '14 at 18:53
  • $\begingroup$ 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. $\endgroup$ Oct 10 '14 at 19:11
  • $\begingroup$ 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. $\endgroup$ 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 $z=\sqrt{2}+1$.

  • $\begingroup$ 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) $\endgroup$
    – A Googler
    Oct 11 '14 at 8:35
  • $\begingroup$ @ A Googler: I have used the fact $|a|-|b|\leq |a+b|$ then because $0\leq |\cos(\alpha)|\leq 1$ the result follows. $\endgroup$
    – Arian
    Oct 11 '14 at 9:46
  • $\begingroup$ Oh yes I got it , fantastic answer! $\endgroup$
    – A Googler
    Oct 11 '14 at 11:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.