Trigo + Complex

Let $A_1,A_2,...A_n$ be a regular polygon of $n$ sides whose centre is origin $O$.Llet the complex numbers representing the vertices $A_1,A_2.....A_n$be$z_1,z_2....z_n$ respectively.The radius is of length unity.Then, \[\prod_{n=2}^{n}|A_1A_n|=\mathbb{?}\]

#ComplexNumbers #Trigonometry #ArgandDiagram

Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
- bulleted - list	bulleted list
1. numbered 2. list	numbered list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
`[example link](https://brilliant.org)`	example link
`> This is a quote`	This is a quote
# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"	# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

Let the circumradius be $R$ . Now we consider all $z$ that satisfy the circle and write a certain expansion:

$z^n-R^n=(z-R)(z^{n-1} + Rz^{n-2}+Rz^{n-3} + \dots + zR^{n-2}+R^{n-1})$

Then we have $z_1=R$ and $,z_2,z_3\dots z_n$ as roots of $(z^{n-1} + Rz^{n-2}+Rz^{n-3} + \dots + zR^{n-2}+R^{n-1})$ , So we can write :

$(z-z_2)(z-z_3) \dots (z-z_n)=z^{n-1} + Rz^{n-2}+Rz^{n-3} + \dots + zR^{n-2}+R^{n-1})$

Since $z_1=R$ is one of the $z$ 's , we substitute $z=R$ in RHS . Thus the equation becomes:

$|(z-z_1)||(z-z_2)||(z-z_3)| \dots |(z-z_n)|=nR^{n-1} \dots (1)$

To calculate $R$ we use cosine rule and get $R=\dfrac{1}{2\sin\left(\dfrac{\pi}{n}\right)}$ . So finally

$\displaystyle\prod_{i=2}^n |A_1A_n|=\dfrac{n}{2\sin^{n-1}\left(\dfrac{\pi}{n}\right)}$

I may be wrong... Adarsh please check

Edit: The problem is edited and the circumradius is unity and not the side. Hence substituting $R=1$ in $(1)$ we get:

$\displaystyle\prod_{i=2}^n |A_1A_n|=n\times 1^{n-1} = \boxed{n}$

Nihar Mahajan - 5 years, 7 months ago

Answer is $n$ .

Adarsh Kumar - 5 years, 7 months ago

Are you sure that the side equals unity? Or the circumradius?

Nihar Mahajan - 5 years, 7 months ago

@Nihar Mahajan – Oh!Sorry!R=1!

Adarsh Kumar - 5 years, 7 months ago

@Adarsh Kumar – Yes , now the answer is $n$ .

Nihar Mahajan - 5 years, 7 months ago

I asked this question a while ago and Alan Yan wrote a nice solution

Otto Bretscher - 5 years, 7 months ago

Thank you for the link!

Nihar Mahajan - 5 years, 7 months ago

Thanks a lot sir!

Adarsh Kumar - 5 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$