Exercise on complex numbers

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Papapetros Vaggelis
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Exercise on complex numbers

#1

Post by Papapetros Vaggelis »

If \(\displaystyle{x\in\mathbb{C}\,,x\neq 0}\) is a complex number such that \(\displaystyle{x+\dfrac{1}{x}=2\,\cos\,a}\)


for some \(\displaystyle{a\in\mathbb{R}}\), then prove that \(\displaystyle{x^{n}+\dfrac{1}{x^{n}}=2\,\cos\,(n\,a)}\)


for every \(\displaystyle{n\in\mathbb{N}}\) .
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Tolaso J Kos
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Re: Exercise on complex numbers

#2

Post by Tolaso J Kos »

Papapetros Vaggelis wrote:If \(\displaystyle{x\in\mathbb{C}\,,x\neq 0}\) is a complex number such that \(\displaystyle{x+\dfrac{1}{x}=2\,\cos\,a}\)


for some \(\displaystyle{a\in\mathbb{R}}\), then prove that \(\displaystyle{x^{n}+\dfrac{1}{x^{n}}=2\,\cos\,(n\,a)}\)


for every \(\displaystyle{n\in\mathbb{N}}\) .
Let $x=\cos a +i \sin a$. Then we immediately see that

$$x+\frac{1}{x}=\cos a +i \sin a + \cos a - i\sin a = 2\cos a$$

Now using De Moivre's formula we have that:

$$x^n +\frac{1}{x^n}=x^n +x^{-n}=\cos na +i\sin na + \cos na - i\sin na =2\cos na$$

as claimed.
Imagination is much more important than knowledge.
Papapetros Vaggelis
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Re:Exercise on complex numbers

#3

Post by Papapetros Vaggelis »

For the complex number \(\displaystyle{x\in\mathbb{C}-\left\{0\right\}}\) holds:

\(\displaystyle{\begin{aligned} x+\dfrac{1}{x}=2\,\cos\,a+i\,0&\implies \overline{x+\dfrac{1}{x}}=\overline{2\,\cos\,a+i\,0}\\&\implies \overline{x}+\dfrac{1}{\overline{x}}=2\,\cos\,a+i\,0\\&\implies \overline{x}+\dfrac{1}{\overline{x}}=x+\dfrac{1}{x}\\&\implies \left(x-\overline{x}\right )-\dfrac{x-\overline{x}}{x\,\overline{x}}=0\\&\implies \left(x-\overline{x} \right )\,\left(1-\dfrac{1}{\left|x\right|^2} \right )=0\\&\implies x=\overline{x}\,\lor \left|x\right|=1\end{aligned}}\)

If \(\displaystyle{x=\overline{x}}\), then \(\displaystyle{x\in\mathbb{R}-\left\{0\right\}}\) and \(\displaystyle{\left|x+\dfrac{1}{x}\right|\geq 2\implies}\)

\(\displaystyle{\implies 2\,\left|\cos\,a\right|\geq 2\implies \left|\cos\,a\right|\geq 1\implies \left|\cos\,a\right|=1\implies \exists\,k\in\mathbb{Z}: a=k\,\pi}\) and

then, \(\displaystyle{\left|x+\dfrac{1}{x}\right|=2\implies x=\pm 1}\) . So, for every \(\displaystyle{n\in\mathbb{N}}\) holds:

\(\displaystyle{x^{n}+\dfrac{1}{x^{n}}=2\,\cos\,(n\,a)}\).

If \(\displaystyle{\left|x\right|=1}\), then \(\displaystyle{\overline{x}=\dfrac{1}{x}}\) and \(\displaystyle{x=\cos\,b+i\,\sin\,b}\) for some \(\displaystyle{b\in\mathbb{R}}\) .

However, \(\displaystyle{x+\dfrac{1}{x}=x+\overline{x}=2\,Re(x)=2\,\cos\,b=2\,\cos\,a\implies x=\cos\,a+i\,\sin\,b}\) and

\(\displaystyle{\left|x\right|^2=1\implies \cos^2\,a+\sin^2\,b=1\implies \sin^2\,b=\sin^2\,a\implies \sin\,b=\pm \sin\,a}\) , so:

for every \(\displaystyle{n\in\mathbb{N}}\) holds :

\(\displaystyle{x^{n}+\dfrac{1}{x^{n}}=x^{n}+\overline{x^{n}}=2\,{\rm Re}(x^{n})=2\,{\rm Re}(\cos(\pm n\,a)+i\,\sin(\pm n\,a))=2\,\cos(\pm n\,a)=2\,\cos\,(n\,a)}\).

There exists a much easier way but this is a more geometrical solution.
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