A better way to find the radius of the complex roots of $(z + 1)^5 = 32z^5$? I was gnawing on this problem today:

All the complex roots of $(z + 1)^5 = 32z^5,$ when plotted in the complex plane, lie on a circle. Find the radius of this circle.

I solved this by first dividing
$$
\left(\frac{z+1}{2z}\right)^5 = 1
$$
then using the roots of unity, and solving for complex $z$. Then, I did regression to determine the solution equation to be
$$
\left(x - \frac{1}{3}\right)^2 + y^2 = \left(\frac{2}{3}\right)^2
$$
Thus, the radius is $2/3$.
Gross. That is (in my opinion) an absolutely awful way to solve this, and not the intended way. I know that there exists a better, non-numerical solution to this problem; Could you please help me find it?
 A: Note that
$$
\frac{z+1}{z}=1+\frac1z
$$
Now solve $$\left(1+\frac1z\right)^5=32$$ with respect to $\frac1z$ and find the circle the solutions lie on. Then apply the reciprocal map to the resulting circle to find the circle where the $z$-solutions lie. In the complex plane, the reciprocal is inversion in the unit circle, composed with complex conjugation.
A: Alternative approach.
The $5$th roots of $32$ are given by $2 \times k$, where
$k$ is one of the $5$ roots of $z^5 = 1.$
These roots are $~\displaystyle e^{i(2k\pi/5)} ~: ~k \in \{0,1,2,3,4\}.$
From the original problem, you have that
$$\left[\frac{z + 1}{z}\right]^5 = 32.$$
Therefore, $(z + 1) = z \times 2 \times e^{i(2k\pi/5)}.$
Therefore $1 = z \times \left[\left(2 \times e^{i(2k\pi/5)}\right) - 1\right].$
Therefore,
$$z = \frac{1}{\left(2 \times e^{i(2k\pi/5)}\right) - 1} = \frac{1}{\left[2\cos(2k\pi/5) - 1\right] + i \left[2\cos(2k\pi/5)\right]}. \tag1$$
Multiplying the numerator and denominator in (1) above by
$\displaystyle \left[2\cos(2k\pi/5) - 1\right] - i \left[2\sin(2k\pi/5)\right]$
gives
$$z = \frac{\left[2\cos(2k\pi/5) - 1\right] - i \left[2\sin(2k\pi/5)\right]}{5 - 4\cos(2k\pi/5)}.$$

Edit
For what it's worth, the $5$ roots of $z^5 - 1 = 0$ may be routinely converted into radical form by noting that
$$z^5 - 1 = (z - 1)(z^4 + z^3 + z^2 + z + 1).\tag2 $$
The 2nd factor in (2) above may be conquered by noting that $z = 0$ will not satisfy the equation.
Therefore, you have that
$$z^2 + z + 1 + \frac{1}{z} + \frac{1}{z^2} = 0.$$
Then, you set $~\displaystyle w = z + \frac{1}{z}$.
Then, you note that $~\displaystyle w^2 = z^2 + \frac{1}{z^2} + 2.$
You end up with a quadratic equation in $w$ which will have $2$ roots, $w_1$ and $w_2$.
You then have $2$ separate quadratic equations :
$$z + \frac{1}{z} = w_1 ~~~\text{and}~~~ z + \frac{1}{z} = w_2.$$
You end up with the $4$ roots of the 2nd factor in (2) above.
