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Find all $z$ such that $(z+1)^7 = z^7.$ You do not have to put your answer in the simplest rectangular or exponential form.

So far I've found that $z^7$ to get $(\frac{z+1}{z})^7 = 1,$ and $\frac{z+1}{z} = e^{\frac{6\pi}{7}}$, $e^{\frac{8\pi}{7}}$, $e^{\frac{10\pi}{7}}$, $e^{\frac{12\pi}{7}}$, $e^{\frac{12\pi}{7}}$, the seven roots of unity. However, I don't know how to proceed.

Any help is greatly appreciated!

Later
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730563
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    you say seven roots, but you write just 5 of them – Exodd Jul 06 '20 at 19:47
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    You have that $\frac{z+1}{z}$=$x_i$, where the $x_i$ are the seven roots of unity. Then you can multiply both sides of the equation by z to get $z+1=zx_i$. Subtract z from both sides to get $1=z(x_i-1)$. Then divide both sides by $x_i-1$ to get $z=\frac{1}{x_i-1}$ to get the seven correct answers – Moko19 Jul 06 '20 at 19:51
  • @exodd Sorry! I meant $1$, $e^{\frac{2\pi}{7}}$, $e^{\frac{4\pi}{7}}$, $e^{\frac{6\pi}{7}}$, $e^{\frac{8\pi}{7}}$, $e^{\frac{10\pi}{7}}$, $e^{\frac{12\pi}{7}}$, $e^{\frac{12\pi}{7}}$. But as Pythagoras explained, (z+1)/z =1 doesn't work. – 730563 Jul 06 '20 at 23:58
  • Hint: $\frac{z+1}{z}=1+\frac{1}{z}$ – A-Level Student Jul 07 '20 at 09:13
  • You have six roots here (the term in $z^7$ cancels). This missing seventh root is a "root at infinity". If you put $w=\frac 1z$ you will find that the equation in $w$ acquires (on clearing fractions) a seventh root $w=0$. – Mark Bennet Sep 30 '20 at 19:57

2 Answers2

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From $(z+1)^7=z^7,$ it is clear that $z\neq 0,-1$. It follows as you did that one needs to solve $$\left(\frac{z+1}z\right)^7=1,$$ hence it seems that $\frac{z+1}z$ has seven choices, namely $1,\zeta,\cdots,\zeta^6$, where $\zeta=e^{\frac{2i\pi}7},$ however the equation $(z+1)/z=1\Leftrightarrow z+1=z$ yields no solutions. To finish up (as you almost did), just solve $$\frac {z+1}z=\zeta^j,j=1,\cdots,6$$ $$\Leftrightarrow z+1=\zeta^jz,j=1,\cdots,6$$ which are linear equations in $z$ with solutions $$z=\frac 1{\zeta^j-1},j=1,\cdots,6.$$ Note that you have $6$ solutions instead of $7$.

Pythagoras
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  • Thanks! Quick question: what does ζ mean? – 730563 Jul 06 '20 at 23:51
  • $\zeta$ is defined in my answer as $\zeta=e^{\frac {2i\pi}7}$, which is a primitive $7$-th root of unity because it generates all other by keeping raising to powers. – Pythagoras Jul 07 '20 at 01:04
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As we can see, both sides of the equation are to the seventh power. We can simplify this equation by dividing by $z^7$ in order to get $(\frac{z+1}{z})^7,$. This works because $0$ is not a solution to this equation. We can see that $\frac{1}{0}$ is undefined.

Since $\frac{z+1}{z}$ can be rewritten as $1+\frac{1}{z}$, we see that this will be a seventh root of unity. If we use the fact that 1 isn't a seventh root of unity ($z+1$ isn't equal to $z$), we see that $0\le k\le 6$, showing 6 possible solutions for $z$. If we use the general form for a seventh root of unity, It's in the form $e^{\frac{2ki\pi}{7}}$ for which $k$ is an integer and $0\le k\le 6$. Since $k$ cant equal $0$, then we will end up with the same problem as before, but if $k = 1$, we see a slightly gross-looking fraction in which we raise $1$ to the negative power, giving us a solution of $\frac{1}{e^{\frac{2i\pi}{7}}-1}$, meaning that the rest of the solutions are in the form $\frac{1}{e^{\frac{2ki\pi}{7}}-1}$. We now have the rest of the solutions for $z$, $\frac{1}{e^{\frac{2i\pi}{7}}-1}$

$\frac{1}{e^{\frac{4i\pi}{7}}-1}$

$\frac{1}{e^{\frac{6i\pi}{7}}-1}$

$\frac{1}{e^{\frac{8i\pi}{7}}-1}$

$\frac{1}{e^{\frac{10i\pi}{7}}-1}$

$\frac{1}{e^{\frac{12i\pi}{7}}-1}$

Peyton Hecht
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