Solutions of the equation $((z-1)/z)^4=1$

Question

The question before this asked to solve $z^4 = 1$ (I found the four roots of 1). I used Euler's formula to solve it and hence solve the question below. I'm just not sure how to go about it exactly.

Solve the equation $\left(\dfrac{z-1}{z}\right)^4 = 1$.

Answer 1

To solve $\left(\frac{z-1}z\right)^4=1$, one can proceed in two steps:

• First, solve $w^4=1$ (apparently, this is already done)
• Then, solve $\frac{z-1}z=w$, for any given $w$ (and then, one is led to treat separately the cases $w=1$ and $w\ne1$)

Answer 2

A somewhat less elegant solution, but no less valid, would be to multiply both sides by $z^4$ , so $(z-1)^4=z^4$, then multiplying the left side out and expanding gives us $z^4-4z^3+6z^2-4z+1=z^4$. Canceling our $z^4$ from either side, gives you $-4z^3+6z^2-4z+1=0$. This is a cubic equation with real coefficients, so the fundemental theorem of algebra says in the complex plane it has exactly 3 (possibly repeated) roots. You can then find the roots with your favorite method, including the cubic formula. However, first you can use the rational roots test to show that if it has a rational root, it's of the form $\pm 1,\pm \frac 1 2 ,\pm \frac 1 4$. Plugging in, we get $\frac 1 2$ is a root, so we can then divide by $z-1/2$ to get the remaining quadratic is $-4z^2+4z-2$, and then use the quadratic formula on that to get the remaining roots of $\frac 1 2 \pm \frac i 2$

Answer 3

Note that $\frac {z-1}{z}=1-\frac 1z $. So you want this expression to be $1,-1, i,-i $. It cannot be $1$, so you are left with three options: $$ z=\frac1 {1-(-1)}=\frac12, \ \ \ z=\frac1 {1-i}=\frac { 1+i} 2, \ \ \ z=\frac1 { 1+i}=\frac { 1-i} 2. $$