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Question

if $S$ be the set of solution of $$\bigg(1+\frac{1}{z}\bigg)^{4}=1$$ then prove that the points are co-linear.

Attempt $\bigg(1+\frac{1}{z}\bigg)^{4}=1$ $\implies z^4+4z^3+6z^2+4z+1=z^4$ $\implies 4z^3+6z^2+4z+1=0$ $\implies (2z+1)(2z^2+2z+1)=0$

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    $\begingroup$ Double-check your solutions to $2z^2 + 2z + 1 = 0$. $\endgroup$ – Connor Harris Apr 16 at 19:36
  • $\begingroup$ Alternatively, just note that the solutions to $2z^2 + 2z + 1 = 0$ are complex conjugates and (by Vieta's formula) have sum $-1$, so the real part of each is $-1/2$. You don't even have to find the imaginary parts that way. $\endgroup$ – Connor Harris Apr 16 at 19:37
  • $\begingroup$ How do you conclude the each have real part $\frac{-1}{2}$ ? Can you elaborate a little. $\endgroup$ – M Desmond Apr 16 at 19:41
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    $\begingroup$ The roots of a polynomial with real coefficients occur in complex conjugate pairs. Therefore, denote the solutions of $2z^2 + 2z + 1 = 0$ by $x + iy$ and $x - iy$. By Vieta's formula, the sum of the solutions of $az^2 + bz + c = 0$ is $-b/a$. Therefore, $2x = -1$. $\endgroup$ – Connor Harris Apr 16 at 19:48
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Just take the fourth root both sides to get $$1+\frac1{z}=\pm1,\pm i$$ $$\frac1{z}=-2,0,-1\pm i$$ $$z=-\frac12,-\frac12\pm \frac12i$$ Which are colinear solutions.

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Hint Simpler soultion:

Let $w=1+\frac{1}{z}$. Then $$w^4=1 \Rightarrow w=1,-1,i,-i$$

$w=1$ is not possible, the other three lead to $$\frac{1}{z}=w-1 \Rightarrow z=\frac{1}{w-1}$$

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$$ \Big((1+{1\over z})-1\Big)\Big((1+{1\over z})+1\Big)\Big((1+{1\over z})+i\Big)\Big((1+{1\over z})-i\Big)=0$$

So $z_1 =-{1\over 2}$ and $\displaystyle z_{2,3} ={-1\pm i\over 2}$

Since $$Re(z_i) = -{1\over 2}$$ for all $i$ the solution are collinear, all the solution are on a line paralle to imaginary axsis.

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