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I'm trying to the solve the equation. Turn out there are two different answers when

$$\left(\frac{z-1}{z+1}\right)^n=1,\left(\frac{z+1}{z-1}\right)^n=1$$

my solution are

$$z=i\cot\left(\frac{k\pi}{n}\right),z=-i\cot\left(\frac{k\pi}{n}\right)$$

My question is which one is correct or they both are correct?

sopanha
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  • Try to solve it explicitly for a couple of small values of $n$ (say, $2$, $3$, and $4$). What actual solutions do you get? What solutions do your cotangents produce? – Arthur Mar 20 '23 at 02:55
  • https://math.stackexchange.com/questions/607487/the-roots-of-the-equation-zn-1zn – lab bhattacharjee Mar 20 '23 at 03:40
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    Since k can be both negative and positive, therefore thy generate the same solution set! – Lai Mar 20 '23 at 03:41
  • I’m voting to close this question because it is simple and was answered in a comment. – Kurt G. Mar 23 '23 at 07:00

1 Answers1

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So basically, both solutions are technically identical. We know that the set of k is basically {1,2,3,... n-1}. Here, we can clearly observe this solution (considering $z = i cot (\frac{kπ}n)$) has two sets of values. For one of the sets, the value of $\frac{k\pi}n$ lies in the 1st quadrant and for the other half it lies in the 2nd quadrant. For example, $\frac{2\pi}n$ is in 1st quadrant while $\frac{(n-2)\pi}n$ is in the second quadrant. Now the cotangents of these 2 angles are equal in magnitude but opposite in sign. In other words, $$cot (\frac{2π}n) = -cot (\frac{(n-2)π}n)$$ Thus, both sets of values are accounted for by either solution! Hence they are both correct but are basically referring to the same thing.

Srish Dutta
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  • This was mentioned in a comment three days ago. That question should be closed, not attract further obvious answers. – Kurt G. Mar 23 '23 at 06:58