Treasure hunt using complex numbers & an inequality

Verdict

Hmm alright. So then is it.. (z1 + z2)(z1* + z2*)? As in, the star is the complex conjugate. If not, I am just not familiar with the formula I guess..

DrClaude

Yes, so you get
[tex]
\left| a + i b \right|^2 = a^2 + b^2
[/tex]

jbunniii

$$\begin{align}
\left|e^{i\theta_1} + e^{i\theta_2}\right|^2
&= \left|e^{i\theta_2} (1 + e^{i(\theta_1 - \theta_2)})\right|^2 \\
&= \left|1 + e^{i(\theta_1 - \theta_2)}\right|^2 \\
&= \left|e^{i(\theta_1 - \theta_2)/2} (e^{-i(\theta_1 - \theta_2)/2} + e^{i(\theta_1 - \theta_2)/2})\right|^2 \\
&= \left| (e^{-i(\theta_1 - \theta_2)/2} + e^{i(\theta_1 - \theta_2)/2})\right|^2 \\
&= \left| 2 \cos((\theta_1 - \theta_2)/2) \right|^2 \\
&= 4 \cos^2((\theta_1 - \theta_2)/2) \\
&= 4 \left(\frac{1}{2} + \frac{1}{2} \cos(\theta_1 - \theta_2)\right) \\
&= 2 + 2 \cos(\theta_1 - \theta_2)\\
\end{align}$$
Probably you can find a way to consolidate a few of the steps to make it shorter.

tiny-tim

Hi Verdict!
Your result is rather difficult to use, because you've chosen to fix the tree even though the tree is variable!

It would be a lot easier to put the two rocks at ±1, and the tree at a general point z

Verdict

Yeah I just finished doing this in a different (longer!) method, but I do indeed end up with that! :) So now I understand how we get there, that's where the tricks come in I guess. But simplifying it further, to what you have suggested, has failed so far...

I wrote out all the multiplications, and you are left with two terms on the left that are smaller than or equal to two terms on the right, which is nice, but a third term on the left that is bigger than or equal to the term on the right. So that isn't as nice as I had hoped it to be, and it is of course not the same as what you reduced it to. Do you use more identities to get there?

@tiny-tim
Hey, thanks! I did manage to work it out in the end (it's somewhere in this heap of posts, haha), and yeah it is a lot harder than it could have been, sadly. Ah well, if it works, it works. The digging twice part is a nuisance, so it is indeed a lesser answer, but the question didn't technically require me to only dig once, so I guess it's not that bad.

jbunniii

Can you show what you currently have? I'll try to talk you through the rest of the simplification.

jbunniii

Here's a much quicker way, FYI:
$$\begin{align}
|e^{i\theta_1} + e^{i\theta_2}|^2
&= \overline{(e^{i\theta_1} + e^{i\theta_2})} (e^{i\theta_1} + e^{i\theta_2}) \\
&= (e^{-i\theta_1} + e^{-i\theta_2}) (e^{i\theta_1} + e^{i\theta_2}) \\
&= 2 + e^{i(\theta_1 - \theta_2)} + e^{-i(\theta_1 - \theta_2)} \\
&= 2 + 2\cos(\theta_1 - \theta_2)
\end{align}$$

Verdict

Alright, so now I have
4r₁r₂ + 2r₁²cos(θ₁-θ₂) + 2r₂²cos(θ₁-θ₂)

is smaller than or equal to 2r₂² + 2r₁² +4 r₁r₂cos(θ₁-θ₂)

So I see that on the LHS the two terms with the cosine are, smaller than or equal to the two corresponding ones on the RHS. However, the cosine on the RHS is smaller than or equal to the term on the left side.


Oh, and one small question:
When you do |z1+z2|², you write (z1+z2)(z1*+z2*). But isn't that just |z1+z2|, without the square?

jbunniii

OK, let's see what we can do with this.
$$\begin{align}
4r_1 r_2 + 2 r_1^2 \cos(\theta_1 - \theta_2) + 2r_2^2\cos(\theta_1 - \theta_2)
\leq 2r_2^2 + 2r_1^2 + 4r_1 r_2 \cos(\theta_1 - \theta_2)\\
\end{align}$$
Moving everything to one side of the inequality, we get
$$\begin{align}
0 &\leq 2r_1^2 (1 - \cos(\theta_1 - \theta_2)) + 2r_2^2 (1 - \cos(\theta_1 - \theta_2)) + 4r_1 r_2 ( \cos(\theta_1 - \theta_2) - 1) \\
&\leq (2r_1^2 + 2r_2^2 - 4r_1 r_2) (1 - \cos(\theta_1 - \theta_2)) \\
\end{align}$$
Now it should be pretty easy to argue that both factors on the right hand side are nonnegative.

Verdict

Alright, so the part with the cosine is obvious indeed: a cosine has a maximal value of 1, so at most, the term is zero, never negative. Inequality holds.
For the first part though.. Of course it makes sense, that if you take something squared, plus something smaller squared, that that is bigger than 2 times something times the smaller one.. It sounds like there is a theorem/inequality for this type of thing, I'll look for it.

Edit: Ok, that was one of the easier things I have missed so far. Got it, inequality proven, thanks so much! You guys are really great.

Also, I made a sneaky edit in my last post:
When you do |z1+z2|², you write (z1+z2)(z1*+z2*). But isn't that just |z1+z2|, without the square?

jbunniii

No. To see that this is not true, consider the special case where both z1 and z2 are real. Then ##(z_1 + z_2)(z_1^* + z_2^*) = (z_1 + z_2)(z_1 + z_2) = (z_1 + z_2)^2 = |z_1 + z_2|^2##, not ##|z_1 + z_2|##.

Verdict

Yeah. Trivial, indeed. Hm, I don't know where the confusion came from, I thought I remembered it from my quantum physics course, but that was clearly not the absolute value.

Again, thanks so much, for all the time and effort!