### Instruction: 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.### Response: I know that the magnitude of a complex number is the distance from the origin in the complex plane, so I want to find the magnitude of the roots of the equation.
I can rewrite the equation as $(z + 1)^5 = 32z^5$ and then divide both sides by $z^5$ to get $\left(\frac{z + 1}{z}\right)^5 = 32.$
Now I can take the fifth root of both sides and get $\frac{z + 1}{z} = \sqrt[5]{32} \cdot e^{2\pi k i / 5},$ where $k$ is an integer from $0$ to $4$.
This means that the roots are of the form $z = \frac{z + 1}{\sqrt[5]{32} \cdot e^{2\pi k i / 5}},$ which I can simplify by multiplying the numerator and denominator by the conjugate of the denominator.
I get $z = \frac{\sqrt[5]{32} \cdot e^{-2\pi k i / 5} (z + 1)}{32}.$
Now I can solve for $z$ and get $z = \frac{\sqrt[5]{32} \cdot e^{-2\pi k i / 5}}{32 - \sqrt[5]{32} \cdot e^{-2\pi k i / 5}}.$
To find the magnitude of $z,$ I need to divide the magnitude of the numerator by the magnitude of the denominator.
The magnitude of the numerator is $\sqrt[5]{32}$ and the magnitude of the denominator is $\sqrt{32^2 + (\sqrt[5]{32})^2}.$
So the magnitude of $z$ is $\frac{\sqrt[5]{32}}{\sqrt{32^2 + (\sqrt[5]{32})^2}}.$
This is the radius of the circle that the roots lie on, and it is independent of $k.$