complex number euler form modulus question (1 Viewer)

gamja · Mar 6, 2023

$prove\ that\ the\ max\ value\ of\ |e^{i\theta}-2|+|e^{i\theta}+2|\ is\ 2\sqrt{5}$

How would you solve this? Apparently there are methods using geometrical graphing or using trig algebra - feel free to solve outside these methods tho.

Much thanks!

yanujw · Mar 6, 2023

The most prominent method is to let $e^{i\theta}=cos(\theta) + isin(\theta)$ . Then use the fact that $|x+iy| = \sqrt{x^2+y^2}$

The locus of z itself is an ellipse which would have been helpful in the old syllabus. However, there is still a geometric way to find the maximum value.

Recall that $|e^{i\theta}-2|+|e^{i\theta}+2|$ means the combined distance of $e^{i\theta}$ from the points 2 and -2 on the Argand diagram. And $e^{i\theta}$ represents the unit circle of points. The combined length of the two vectors as visualised below is maximised* at either the points i or -i (see diagram below). At these points, each line has a length of $\sqrt{5}$ so the maximum combined length is $2\sqrt{5}$

*this is just a vague conclusion based on looking at the graph, it may require an actual proof which just goes back to the algebraic version anyways.

tywebb · Mar 6, 2023

You can also do it with calculus.

$f(\theta)=|e^{i\theta}-2|+|e^{i\theta}+2|\text{ can be simplified to }\sqrt{5-4\cos\theta}+\sqrt{5+4\cos\theta}$

$\text{Then }f'(\theta)=\frac{2\sin\theta}{\sqrt{5-4\cos\theta}}-\frac{2\sin\theta}{\sqrt{5+4\cos\theta}}$

$\text{Maximum value occurs when }f'(\theta)=0\text{ so }\sin\theta=0\text{ or }\cos\theta=0$

$\sin\theta=0\text{ actually gives the minimum value } 4 \text{ and }\cos\theta=0\text{ gives the maximum value }2\sqrt5$

$\text{Checking w.l.o.g. maximum value occurs if }\theta=\frac{\pi}{2}\text{ and }f(\frac{\pi}{2})=2\sqrt5$

$\text{You can see from the graph of }y=f(\theta)\text { that this maximum value }y=2\sqrt5\text{ also occurs}$

$\text{for all odd multiples of }\frac{\pi}{2}\text{ i.e., at }\theta=\frac{(2k+1)\pi}{2}\ \forall k\in\Bbb Z$

Drongoski · Mar 6, 2023

yanujw said:
The most prominent method is to let $e^{i\theta}=cos(\theta) + isin(\theta)$ . Then use the fact that $|x+iy| = \sqrt{x^2+y^2}$

The locus of z itself is an ellipse which would have been helpful in the old syllabus. However, there is still a geometric way to find the maximum value.

Recall that $|e^{i\theta}-2|+|e^{i\theta}-2|$ means the combined distance of $e^{i\theta}$ from the points 2 and -2 on the Argand diagram. And $e^{i\theta}$ represents the unit circle of points. The combined length of the two vectors as visualised below is maximised* at either the points i or -i (see diagram below). At these points, each line has a length of $\sqrt{5}$ so the maximum combined length is $2\sqrt{5}$

*this is just a vague conclusion based on looking at the graph, it may require an actual proof which just goes back to the algebraic version anyways.

View attachment 37914

The geometric method (see your diagram) is the easiest to understand and the shortest.

tywebb · Mar 8, 2023

Some dude (can't remember who) came here before and tried to use the AM-GM inequality to do it but it didn't work. They kind of had a good idea but had the inequality the wrong way around.

However the AM-QM inequality does work (instead of the AM-GM one):

$\begin{aligned}|e^{i\theta}-2|+|e^{i\theta}+2|&\le2\sqrt{\frac{|e^{i\theta}-2|^2+|e^{i\theta}+2|^2}{2}}\\&=2\sqrt{\frac{(\cos\theta-2)^2+\sin^2\theta+(\cos\theta+2)^2+\sin^2\theta}{2}}\\&=2\sqrt{\frac{\cos^2\theta+\sin^2\theta-4\cos\theta+4+\cos^2\theta+\sin^2\theta+4\cos\theta+4}{2}}\\&=2\sqrt{\frac{10}{2}}\\&=2\sqrt5\text{ with equality }\iff|e^{i\theta}-2|=|e^{i\theta}+2|\iff\cos\theta=0\iff\theta=\frac{(2k+1)\pi}{2}\forall k\in\Bbb Z\end{aligned}$

member 6003 · Mar 8, 2023

tywebb said:
Some dude (can't remember who) came here before and tried to use the AM-GM inequality to do it but it didn't work. They kind of had a good idea but had the inequality the wrong way around.

However the AM-QM inequality does work (instead of the AM-GM one):

$\begin{aligned}|e^{i\theta}-2|+|e^{i\theta}+2|&\le2\sqrt{\frac{|e^{i\theta}-2|^2+|e^{i\theta}+2|^2}{2}}\\&=2\sqrt{\frac{(\cos\theta-2)^2+\sin^2\theta+(\cos\theta+2)^2+\sin^2\theta}{2}}\\&=2\sqrt{\frac{\cos^2\theta+\sin^2\theta-4\cos\theta+4+\cos^2\theta+\sin^2\theta+4\cos\theta+4}{2}}\\&=2\sqrt{\frac{10}{2}}\\&=2\sqrt5\text{ with equality }\iff|e^{i\theta}-2|=|e^{i\theta}+2|\iff\cos\theta=0\iff\theta=\frac{(2k+1)\pi}{2}\forall k\in\Bbb Z\end{aligned}$

I was the dude, cool solution but ofc its outside the syllabus. I saw 2sqrt(5) in my solution so I just thought it would be correct without considering any logic before posting.

tywebb · Mar 8, 2023

I don't think it is outside the syllabus if you prove this first (pretty easy to do):

$\text{Prove that for all non-negative real numbers }a,b,\frac{a+b}{2}\le\sqrt{\frac{a^2+b^2}{2}}\text{ with equality }\iff a=b$

member 6003 · Mar 8, 2023

well if you prove first then yea they couldn't mark you down for it

complex number euler form modulus question (1 Viewer)

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