Sum of angles in regular polygon

By Arun Iyer on Monday, January 21, 2002 - 06:58 pm:

this problem is from some book on complex number and was posed before me by my friend...the problem really seems out of this world atleast for me...

the problem.....
Given->
ABCDEFG..... is a regular polygon of n sides,inscribed in a circle of radius a and center O.

Show that->
the sum of the angles that AP,BP,CP,.... make with OP is,
$\tan^{- 1} ([a^{n} \sin n θ] / [a^{n} \cos n θ - r^{n}])$

where OP $= r$ and angle AOP $= θ$
love arun

By David Loeffler on Thursday, January 24, 2002 - 01:13 pm:

My reasoning was as follows. Let's suppose we are in the complex plane and

O

is the origin, and

A

the point

(a, 0)

. Then the other vertices are the solutions of the equation

x^{n} = a^{n}

, so we obtain

$x^{n} - a^{n} = (x - a) (x - b) (x - c) \dots$

Now, the angle you want is the sum of the arguments of $(a - x) - \arg x$ , $(b - x) - \arg x$ etc where $x$ is the point $P$ .

So this is

$\arg (a - x) (b - x) \dots - \arg x^{n} = \arg (- 1)^{n} (x - a) (x - b) \dots - \arg x^{n} = \arg (- 1)^{n} (1 - (a / x)^{n})$

Now put $x = r e^{i θ}$ , so $x^{n} = r^{n} e^{n i θ}$ , and we must find

$\arg (- 1)^{n} (1 - a^{n} r^{- n} e^{- n i θ})$

$= \tan^{- 1} [(a^{n} \sin n θ) / (a^{n} \cos n θ - r^{n})]$

(possibly $\pm$ some multiple of $π$ ).

David