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Let $z$ and $z+1$ be complex numbers such that $z$ and $z+1$ are both $n^{\text{th}}$ complex roots of $1$. If $n$ is a multiple of $5$, compute the minimum value of $n+z^3$.

What I started out with was $(\operatorname{cis}\theta)^n=(\operatorname{cis}\theta+1)^n=1$.

Simplifying the right side, I got that $(\operatorname{cis}\theta)^n=\left(\sqrt{2+2\cos\theta}\operatorname{cis}\frac{\theta}{2}\right)^n=1$. I did not know what to do next so I decided to equate the moduli to see what I got. I got that $\theta=\frac{2\pi}{3}$ or $\theta=\frac{4\pi}{3}$. However, I do not know what to do with those values or even if I am doing the correct thing.

Any advice would be appreciated.

2 Answers2

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It's ok. Note that $z^3=1$ in both cases (because $z^3$ has module 1 and argument $2\pi$ or $4\pi$).

This means that $3\mid n$. But if $z+1$ is nth root, then $2\mid n$

Then $n+z^3$ is constant $n+1$

Now, the min of $n$ is 2.3.5=30 and we have that 31 is the answer

Martín Vacas Vignolo
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  • I don't get the part where $2\mid n$. Why is that true? –  May 03 '18 at 03:22
  • @silverpegasus258 because $z$ and $z+1$ are simmetric respect to the axis y. Think that nth roots are the vertices of n-regular-polygon – Martín Vacas Vignolo May 03 '18 at 03:35
  • @silverpegasus258 -- We know that $z = -\frac12\pm\frac{\sqrt3}{2}i$, so $z+1 = \frac12\pm\frac{\sqrt3}{2}i$, which means that $(z+1)^6 = 1$, but $(z+1)^3 = -1 \neq 1$. So $n=3$ is not a possibility; $n$ must be a multiple of $6$. – mr_e_man May 03 '18 at 03:37
  • In other words, the triangle with vertices in $z$, $z+1$ and $0$ is equilateral (because all sides are 1) and then $z+1$ has argument $\pm\pi/3$ ie, exist k such that $\pm 1/3=2k/n$ and then $n$ is even. – Martín Vacas Vignolo May 03 '18 at 03:40
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A different way to find the candidates for $z$ and $z+1$ is to note that they are a distance apart of one on a horizontal line yet they both have to lay on the unit circle. It just so happens that fitting an equilateral triangle with one vertex in the center of the circle at the other two on the circle fits the bill.

This makes the z values either one third around the circle or two thirds around the circle as you have discovered. The two possible $z$ values are cube roots of unity. The corresponding $z+1$ values are sixth roots of unity. Cube roots are automatically sixth roots so all four are sixth roots.

In a similar manner, they are also $n^{th}$ roots when $n$ is a multiple of 6.

That, along with what MVV said, should let you easily solve this.

Hope this helps.

Ced

Cedron Dawg
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