How do I solve $(x+1)^n=(x-1)^n$? I assumed $x=a+bi$, getting the equation $((a+1)+bi)^n=((a-1)+bi)^n$. How do I solve it using Moivre's n-th root theorem?
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3 Answers
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Hint: $x = 1$ is not a solution, so we might assume $x \neq 1$ and re-write this as:
$$\left(\frac{x + 1}{x - 1} \right)^n = 1$$
It boils down to just listing the $n$th roots of $1$ and computing $x$.
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$\begingroup$ Why can I equal this equation to 1? $\endgroup$ Jul 2, 2015 at 18:03
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1$\begingroup$ @FrancieleDaltoé: by diving both sides of the initial equation by $(x -1)^n$. $\endgroup$– user230734Jul 2, 2015 at 18:03
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Your equation is equivalent to $$ \left(\frac{x+1}{x-1}\right)^n = 1 \tag{1}$$ and since the inverse function of $f(x)=\frac{x+1}{x-1}$ is given by $\frac{y-1}{y+1}$, we have: $$ x = \frac{\exp\frac{2\pi k i}{n}+1}{\exp\frac{2\pi k i}{n}-1}=-i\cot\frac{\pi k}{n},\qquad k\in\left[1,\ldots,n-1\right].\tag{2}$$
answered Jul 2, 2015 at 18:02
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$\begingroup$ There seems to be a sign error for odd $n$. For example, for $n=3$, one root given by $(2)$ is $x=i\tan(\pi/3)$. Then, $\frac{x+1}{x-1}=\frac{i\tan(\pi/3)+1}{i\tan(\pi/3)-1}=\frac{\cos(\pi/3)+i\sin(\pi/3)}{-\cos(\pi/3)+i\sin(\pi/3)}=-e^{2\pi/3}$ and $\left(-e^{2\pi/3}\right)^3=-1\ne 1$. $\endgroup$ Jul 2, 2015 at 20:00
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$\begingroup$ @Dr.MV: you are right, now fixed, thanks. $\endgroup$ Jul 2, 2015 at 20:02
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$\begingroup$ Glad the comment was useful! It is interesting that there were 4 up votes before the edit. ;-)) $\endgroup$ Jul 2, 2015 at 20:03
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$\begingroup$ @FrancieleDaltoé: so you have to solve $y^n=1$, then $\frac{z+1}{z-1}=y$, that is exactly what I did. $\endgroup$ Jul 3, 2015 at 14:53
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$\begingroup$ Is there any simples way to solve it? I need to use basic concepts of complex numbers. I wrote: $(z+1)^n=(z-1)^n \leftrightarrow \frac{(z+1)^n}{(z-1)^n}=\frac{(z-1)^n}{(z-1)^n}=1 \leftrightarrow \left (\frac{z+1}{z-1}\right)^n=1 \leftrightarrow \sqrt[n]{\left (\frac{z+1}{z-1}\right)^n}=\sqrt[n]{1}$ What do I do from there? I thought I'd write $\frac{z+1}{z-1}=x$ and then solve $x^n=1=(\cos(2k\pi)+i\sin(2k\pi))$, but I'm not sure how to do that. $\endgroup$ Jul 3, 2015 at 14:58
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For $z^n=w$, one can write $w=we^{i2 \pi \ell}$ for integer $\ell$. Then, solving for $z$ reveals that
$$z=w^{1/n}e^{i2 \pi \ell/n}$$
for $\ell =0, 1, 2,\cdots n-1.$
Now for $z=x+1$ and $w=(x-1)^n$ we have that
$$x+1=(x-1)e^{i2 \pi \ell/n} \implies x = \frac{e^{i2 \pi \ell/n}+1}{e^{i2 \pi \ell/n}-1}=-i\cot{\pi\ell/n}\,\,\text{for}\,\,\ell =1,2,\cdots n-1$$
answered Jul 2, 2015 at 18:48