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Question

Prove that all roots of $(z+1)^n = z^n$ lies on a straight line given that $n$ is a natural numbers for all $n \ge 2$.

Should i start the question assuming $z$ to be of the form

$z= x+iy$ or

$z=re^i\theta$

or the polar form

Em.
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Nitro phenol
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2 Answers2

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One approach: $$(z+1)^n=z^n\implies\vert z+1\vert^n=\vert z\vert^n\implies\vert z+1\vert=\vert z \vert$$

and this is the line $\Re z = -\frac{1}{2}$.

(edited to remove unnecessary passage to fractions)

πr8
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  • oh this implies they lie on aperpendicular bisector of the line joining 0 and -1 that is x= -1/2 right – Nitro phenol Jan 20 '16 at 06:39
  • Correct, yes - if we rewrite it as $\vert z - (-1) \vert = \vert z - 0\vert$, this may be clearer. – πr8 Jan 20 '16 at 06:40
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hint

Note that $z \neq 0$, so we can rewrite the given equation as $$\left(1+\frac{1}{z}\right)^n=1$$ Thus $1+\frac{1}{z}$ should be a $n^{\text{th}}$ root of unity.

Anurag A
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  • how does that prove that all roots lie on a line according to you the points would lie on a circle of unit radius – Nitro phenol Jan 20 '16 at 06:37
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    @PrateekChauhan you need to solve for $z$ and then you get that they lie on the straight line. This is meant to be a hint, not a solution. – Anurag A Jan 20 '16 at 06:42
  • @PrateekChauhan actually $\pi r8$ has given the complete answer. – Anurag A Jan 20 '16 at 06:46
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    This is a fair hint - you can use it to show that $1+\frac{1}{z}=e^{\frac{2\pi i k}{n}}$, which gives $z=\frac{1}{e^{\frac{2\pi i k}{n}}-1}=-\frac{1}{2}-\frac{1}{2}i\cot\frac{k\pi}{n}$ (after a bit of work), which clearly lie on a straight line. – πr8 Jan 20 '16 at 06:57
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    @PrateekChauhan I think trying to ridicule is not a right approach to asking questions ad learning mathematics. Moreover if you cannot complete a hint you can ask for more help instead of rushing to the judgement of whether one can solve it or not. There are multiple ways to proceed from the hint I gave, one way is to talk about the magnitude of the complex number the way $\pi r8$ did it (once he gave the answer there is no reason left for me to redo it), other way is to solve for $z$. – Anurag A Jan 20 '16 at 06:57
  • im srry @ Anurag A i realised that i just spoke too much and after jst the comment i tried the hint and reached the result sorry for being overexpressive and impulsive and thnx for the help both of you @$\pi$r8 and @Anurag A – Nitro phenol Jan 20 '16 at 07:00