1

Let $u, v \in \Bbb C \setminus \{0\}$ such that $u^{2n} + v^{2n} − u^n v^n = 0$ where $n \in \Bbb N^+$.

(a) Show that $$u = v \left(\cos\left(\frac{6k+1}{3n}\pi\right)\pm i\sin\left(\frac{6k+1}{3n}\pi\right)\right)$$ for $k=0,1,2,\ldots, n-1.$

(b)

i) Using (a), solve the equation $(z+ai)^{2n}+(z-ai)^{2n} - (z^2+a^2)^n=0$ where a is a non-zero real number.

ii) Deduce that $$\prod\limits_{k=1}^{n-1}\cot^2\left(\frac{6k+1}{6n}\pi\right) = 2 - (-1)^n$$

For part (a), we can solve the original equation by regarding it as a quadratic equation with the variable u and we have

$$u^n=\frac{v^n±\sqrt{v^{2n}-4v^{2n}}}2=(\frac{1}2-\frac{\sqrt{3}}2i)v^n$$

Hence $u=\sqrt{(\frac{1}2)^2+(\frac{\sqrt{3}}{2})^2} e^{i(\frac{5\pi}{3}+\frac{2k\pi}{n})}v=(\cos(\frac{6k+1}{3n}\pi)±i\sin(\frac{6k+1}{3n}\pi))v$ for $k=0,1,2,\ldots, n-1.$

Therefore

$$u=v(\cos(\frac{6k+1}{3n}\pi)±i\sin(\frac{6k+1}{3n}\pi))$$

for $k=0,1,2,\ldots, n-1$. For part (b) (i), results holds by simply putting $u=z+ai$ and $v=z-ai$

$z=\frac{(\cos(\frac{6k+1}{3n}\pi)-i\sin(\frac{6k+1}{3n}\pi)+1)ai}{\cos(\frac{6k+1}{3n}\pi)-i\sin(\frac{6k+1}{3n}\pi)+1}$

However, I am quite stuck in the last part.

Rodrigo de Azevedo
  • 23,223
Tomas Wang
  • 39

1 Answers1

0

Here is an answer that I have got. However, it is indeed more complicated than I expected, and it requires a property from one of my previous questions. See $$z^{2n} - 2 a^n z^n \cos (nθ) + a^{2n}= \prod_{k=0}^{n-1}\left[z^2-2az\cos\left(\theta+\frac{2\pi k}{n}\right)+a\right]$$

for any positive integer a.

Now, putting $a=1, \cosθ=1/2$, i.e. $θ=\frac{\pi}3$, we have

$$z^{2n}-z^n+1=\prod\limits_{k=1}^{n-1}[z^2-2z\cos(\frac{\pi}{3}+\frac{2k\pi}{n})+1]=\prod\limits_{k=1}^{n-1}[z^2-2z\cos(\frac{6k+1}{3n}\pi)+1]$$

Now, take $z=1$, we have

$$\prod\limits_{k=1}^{n-1}[2-2\cos(\frac{6k+1}{3n}\pi)]=2^{n}\prod\limits_{k=1}^{n-1}[1-\cos(\frac{6k+1}{3n}\pi)]=2^{n}\prod\limits_{k=1}^{n-1}2\sin^2(\frac{6k+1}{6n}\pi)$$

$$=1 \ldots\ldots(*)$$

Similarly, take $z=-1$, we have

$$\prod\limits_{k=1}^{n-1}[2+2\cos(\frac{6k+1}{3n}\pi)]=2^{n}\prod\limits_{k=1}^{n-1}[1+\cos(\frac{6k+1}{3n}\pi)]=2^{n}\prod\limits_{k=1}^{n-1}2\cos^2(\frac{6k+1}{6n}\pi)$$

$$=2-(-1)^n\ldots\ldots(**)$$

Using $\frac{(**)}{(*)}$ we have

$$\frac{2^{n}\prod\limits_{k=1}^{n-1}2\cos^2(\frac{6k+1}{6n}\pi)}{2^{n}\prod\limits_{k=1}^{n-1}2\sin^2(\frac{6k+1}{6n}\pi)}=\prod\limits_{k=1}^{n-1}\frac{\cos^2(\frac{6k+1}{6n}\pi)}{\sin^2(\frac{6k+1}{6n}\pi)}=\prod\limits_{k=1}^{n-1}\cot^2({\frac{6k+1}{6n}\pi})=2-(-1)^n$$ as desired.

Tomas Wang
  • 39