Question

The complex numbers $z_1, z_2 \dots z_n$, represents the vertices of a regular polygon of n sides in order, inscribed in a circle of unit radius and $z_3 + z_n = Az_1 + \overline{A}z_2$, which of the following statement/s is/are correct

• A. if n = 4 then $A = \sqrt{5}$
• B. if n = 6 then $A = \sqrt{4}$
• C. if n = 8 then $A = \sqrt{3}$
• D. if n = 12 then $A = \sqrt{2}$

Answer 1

None of the options are correct.

Answer 2

We shall show that if the vertices of a regular n–gon (with unit circumradius) are taken as

$z_k=e^{2\pi i (k-1)/n}\quad (k=1,2,\ldots n)$,

then $z_3+z_n=e^{4\pi i/n}+e^{-2\pi i/n}$

and

$z_1=1,\quad z_2=e^{2\pi i/n}$.

The equation to be satisfied is $z_3+z_n=A\,z_1+\overline{A}\,z_2\quad\Longrightarrow\quad e^{4\pi i/n}+e^{-2\pi i/n}=A+\,\overline{A}\,e^{2\pi i/n}\,.$

A common way to “solve” an equation like this is to try to make a guess that $A$ is a (real) constant (that is, $A=\overline{A}$). (Indeed, in many such problems one eventually finds a symmetry forcing a real answer.) Assuming $A\in\Bbb R$,

the equation becomes $e^{4\pi i/n}+e^{-2\pi i/n}=A\Bigl(1+e^{2\pi i/n}\Bigr).$

Now equate the imaginary and real parts. Write $e^{4\pi i/n}=\cos\frac{4\pi}{n}+i\sin\frac{4\pi}{n},\quad e^{2\pi i/n}=\cos\frac{2\pi}{n}+i\sin\frac{2\pi}{n},\quad e^{-2\pi i/n}=\cos\frac{2\pi}{n}-i\sin\frac{2\pi}{n}\,.$

Then

$e^{4\pi i/n}+e^{-2\pi i/n}=\Bigl[\cos\frac{4\pi}{n}+\cos\frac{2\pi}{n}\Bigr]+i\Bigl[\sin\frac{4\pi}{n}-\sin\frac{2\pi}{n}\Bigr]$

and

$A\Bigl(1+e^{2\pi i/n}\Bigr)=A\Bigl[1+\cos\frac{2\pi}{n}\Bigr]+ i\,A\sin\frac{2\pi}{n}\,.$

Thus equating imaginary parts we need $\sin\frac{4\pi}{n}-\sin\frac{2\pi}{n} = A\,\sin\frac{2\pi}{n}\,.$

Provided that $\sin\frac{2\pi}{n}\neq 0$ (which is the case for $n\ge4$) we get $A=\frac{\sin(4\pi/n)-\sin(2\pi/n)}{\sin(2\pi/n)}\,.$

Using the sine subtraction formula: $\sin\frac{4\pi}{n}-\sin\frac{2\pi}{n} = 2\cos\frac{3\pi}{n}\sin\frac{\pi}{n}\,,$

and writing $\sin\frac{2\pi}{n}=2\sin\frac{\pi}{n}\cos\frac{\pi}{n}\,,$

we obtain $A=\frac{2\cos\frac{3\pi}{n}\sin\frac{\pi}{n}}{2\sin\frac{\pi}{n}\cos\frac{\pi}{n}}=\frac{\cos\frac{3\pi}{n}}{\cos\frac{\pi}{n}}\,.$

Thus, if $A$ is assumed real then $A=\frac{\cos(3\pi/n)}{\cos(\pi/n)}\,.$

Let us check each option:

1. For $n=4$: $A=\frac{\cos(3\pi/4)}{\cos(\pi/4)}=\frac{-\frac{\sqrt2}{2}}{\frac{\sqrt2}{2}}=-1.$

Option (A) claims $A=\sqrt5$; false.

2. For $n=6$: $A=\frac{\cos(3\pi/6)}{\cos(\pi/6)}=\frac{\cos(\pi/2)}{\cos(\pi/6)}=\frac{0}{\sqrt3/2}=0.$

Option (B) claims $A=\sqrt{4}=2$; false.

3. For $n=8$: $A=\frac{\cos(3\pi/8)}{\cos(\pi/8)}.$

Numerically, $\cos(3\pi/8)\approx0.3827$ and $\cos(\pi/8)\approx0.9239$ so that $A\approx0.414.$

Option (C) claims $A=\sqrt{3}\approx1.732$; false.

4. For $n=12$: $A=\frac{\cos(3\pi/12)}{\cos(\pi/12)}=\frac{\cos(\pi/4)}{\cos(\pi/12)}=\frac{\frac{\sqrt2}{2}}{\cos(\pi/12)}.$

Numerically, $\cos(\pi/12)\approx0.9659$ so that $A\approx\frac{0.7071}{0.9659}\approx0.732,$

whereas Option (D) claims $A=\sqrt2\approx1.414$; false.

Thus none of the given statements (A)–(D) is correct.

In summary:

• None of the options are correct. The calculated value of A does not match any of the options provided for n = 4, 6, 8, or 12.