Solve the equation
$$((z-1)^{20}-2)((z+2)^{10}+1024) = 0$$
and plot the solutions in the complex plane.
I'd like plot to be electronically generated but it shouldn't be too hard to come up with the solution by hand and provide a concrete explanation as to why the roots lie where they are.
To begin solving this equation, we can treat it as if it has already been factored. This way, we can set each factor equal to zero independently and solve for $z$. Setting the first factor equal to zero results in the equation $(z-1)^{20}-2=0$. Therefore, $z=\pm2^{1/20}+1$. However, this only gives us two solutions and we expect $20$. This is because the remaining $18$ solutions lie in the complex plane. To reach them, we must rewrite our equation as $$\begin{align}\pm2^{1/20}+1&=\pm(2\cdot1)^{1/20}+1\\&=\pm(2e^{2k\pi i})^{1/20}+1\\&=2^{1/20}(\pm e^{k\pi i/10})+1\end{align}$$ $\forall \ (k\in\mathbb{Z})\ni(0\leq k\leq20)$, though the solutions corresponding to $k=0$ and $k=20$ share the same value. Because $\theta$ indicates the orientation of a complex number $re^{i\theta}$, including only one exponential function or multiple functions with the same arguments
two exponential functions with different values for $\theta$ are needed to span the whole complex plane.
We can drop the $\pm$ since for each $+e^{k\pi i/10}$ the complex conjugate $+e^{-k\pi i/10}$ is also a solution.
Proof. Let $K$ be the set of integers $k$ such that $0\leq k\leq20$. For a given $k\in K$, $$\begin{align}e^{-k\pi i/10}&=e^{(-k\pi/10+2\pi)i}\\&=e^{(-k\pi/10+20\pi/10)i}\\&=e^{(20-k)\pi i/10}\end{align}$$ Since $0\leq(20-k)\leq20$, $(20-k)\in K$. Therefore, $e^{-k\pi i/10}$ can be expressed in terms of a member of $K$. QED.
Solutions are of the form $z_k=2^{1/20}e^{k\pi i/10}+1 \ \forall \ (k\in\mathbb{Z})\ni(0\leq k\leq20)$. To make these solutions unique, we allow $0\leq k<20$. Additionally, the equation $(z-1)^{20}-2=0$ can be rewritten $|z-1|=2^{1/20}$. This defines a circle of radius $2^{1/20}$ centered at $z=1$ in the complex plane, and indeed, all solutions lie equidistantly along this curve.
The second set of solutions is derived from the equation $(z+2)^{10}+1024=0$. Simplifying gives $z=\pm2(-1)^{1/10}-2$. This can be rewritten $$\begin{align}\pm2(-1)^{1/10}-2&=\pm2(-1)^{1/10}-2\\&=\pm2(e^{(2k+1)\pi i})^{1/10}-2\\&=2(\pm e^{(2k+1)\pi i/10})-2\end{align}$$ $\forall \ (k\in\mathbb{Z})\ni(0\leq k\leq10)$, though the solutions corresponding to $k=0$ and $k=10$ share the same value. We can drop the $\pm$ since for each $+e^{(2k+1)\pi i/10}$ the complex conjugate $+e^{-(2k+1)\pi i/10}$ is also a solution (the proof is analogous to that for the first set of solutions).
Solutions are of the form $z_k=2e^{(2k+1)\pi i/10}-2 \ \forall \ (k\in\mathbb{Z})\ni(0\leq k\leq10)$. To make these solutions unique, we allow $0\leq k<10$. Additionally, the equation $(z+2)^{10}+1024=0$ can be rewritten $|z+2|=2$. This defines a circle of radius $2$ centered at $z=-2$ in the complex plane, and indeed, all solutions lie equidistantly along this curve.
Graph of the complex plain with solutions. Solutions to $(z-1)^{20}-2=0$ form the circle to the right and solutions to $(z+2)^{10}+1024=0$ form the circle the left.
I got bored and wanted to post, so I solved and plotted this problem using Python:
here is the solution 
%matplotlib inline
#import scipy for root solver, matplotlib for plots,
#and sympy to generate coefficient vector
import scipy as sp
from matplotlib import pyplot as plt
from sympy import symbols
#define z as a variable
z=symbols('z')
#define or function of z
f=((z-1)**20-2)*((z+2)**10+1024)
#generate coefficient vector
coeffs = f.as_poly().all_coeffs()
#roots solves and outputs paired vectors of real and imaginary parts
roots=sp.roots(coeffs)
#define our plot
im,ax=plt.subplots()
#put the roots on as 'ro' which are red dots, the x cords are the real parts
#and the y cords are the imaginary parts
ax.plot(roots.real,roots.imag,'ro')
#add the grid on both axes
ax.grid(True, which='both')
#horizontal axis at y=0
ax.axhline(y=0, color='k')
#vertical axis at x=0
ax.axvline(x=0, color='k')Here's the resulting plot:
I used this and this to figure out how to get the coefficients into the right format using sympy and then solve using scipy.roots.
My system clock is off so it says I posted this 4 hours ago.
My clock says you posted around 9:00 on Friday night. Not sure if that's impressive studiousness or just weird. 