Find $\theta$ and $\phi$ that maximize $\mid -2ia\sin\theta - 2ib\sin\phi + 2c(1-\cos\theta) +2d(1-\cos\phi)\mid$

Question

How can you find for what values of the $\theta$ and $\phi$ angles the following modulus will assume its greatest possible value?

$$\mid -2ia\sin(\theta) - 2ib\sin(\phi) + 2c(1-\cos(\theta)) +2d(1-\cos(\phi))\mid$$

$i$ is complex, $a$, $b$, $c$, and $d$ are free parameters. I want to find for what $\theta$ and $\phi$ this expression maximizes because I want to find the values the parameters can take to keep the modulus at less than $1$.

Any help is much appreciated.

Answer 1

Let $z_{1}=x_{1}+y_{1}i=c(1-\cos \theta)-ai\sin \theta$, then

$$\frac{(x_{1}-c)^2}{c^2}+\frac{y_{1}^2}{a^2}=\frac{c^2\cos^2 \theta}{c^2}+\frac{a^2\sin^2 \theta}{a^2}=1$$

Let $z_{2}=x_{2}+y_{2}i=-d(1-\cos \phi)+bi\sin \phi$, then $$\frac{(x_{2}+d)^2}{d^2}+\frac{y_{2}^2}{b^2}=\frac{d^2\cos^2 \theta}{d^2}+\frac{b^2\sin^2 \theta}{b^2}=1$$

Now your expression becomes $2|z_{1}-z_{2}|$.

Up to scaling, it's equivalent to find the maximal distance between the ellipses:

$$ \left \{ \begin{array}{rcl} \displaystyle \frac{(x-c)^2}{c^2}+\frac{y^2}{a^2} &=& 1 \\ \displaystyle \frac{(x+d)^2}{d^2}+\frac{y^2}{b^2} &=& 1 \end{array} \right.$$

Considering their tangents with common slope $m$:

$$ \left \{ \begin{array}{rcl} y &= &m(x-c) \pm \sqrt{c^2 m^2+a^2} \\ y &= &m(x+d) \mp \sqrt{d^2 m^2+b^2} \end{array} \right.$$

The distance between the two tangents is:

$$p_{\pm}=\left| \frac{\pm \sqrt{c^2 m^2+a^2}-cm \pm \sqrt{d^2 m^2+b^2}-dm} {\sqrt{1+m^2}} \right|$$

We can maximize $p_{\pm}$ graphically or numerically when $a,b,c,d$ are specified.

Note that $p_{+}$ and $p_{-}$ are mirror images, just pick up the absolute maximum of $2p_{+}$.