<s>[INST] Solving $(z+1)^5 = z^5$ The question says to solve this equation: $(z+1)^5 = z^5$
I did. Just want to find out if I did it properly and if my run-around logic makes sense.
First I begin my writing the equations as:
$$ (z+1)^5 = z^5$$
$$ \mathbf{e}^{5 \mathbf{Log}(z+1)} = \mathbf{e}^{5 \mathbf{Log}(z)} $$
So $$  \mathbf{Log}(z+1) = \ln|z+1| + \mathbf{Arg}(z+1)i $$ 
 $$  \mathbf{Log}(z) = \ln|z| + \mathbf{Arg}(z)i $$ 
Now, because the natural logarithm is one-to-one, I write:
$$ \ln |z| = \ln |z+1| \Rightarrow |z| = |z+1|$$ 
So assign $ z = a +bi$ 
So that $|z| = \sqrt{a^2 +b^2} = |z+1| = \sqrt{(a+1)^2 +b^2} \Rightarrow a^2 +b^2 = (a+1)^2 +b^2 \Rightarrow a^2 = (a+1)^2 \Rightarrow a = -\frac12 $
So, $z = -\frac12 + bi$ and $z+1 = \frac12 + bi$ for some $b \in \mathbb R$
Now to find $b$
$$ \mathbf{Arg}(z+1) = \mathbf{Arg}(z)$$
$$ \tan^{-1} \frac{b}{\frac12} = \pi - \tan^{-1}\frac{b}{-\frac12}$$
I have a feeling this last part isn't quite right, so I just want to find out if I'm approaching this question properly?
Ultimately, I get $ z = -\frac12$ which upon inspection...is wrong...
 [/INST] If you divide both sides by $z^5$ (note that $z\ne0$, since for $z=0$ we get $0=1$), you get
$$
\left(1+\frac1z\right)^5=1.
$$
The expression in brackets cannot be $1$, so we are left with the four non-trivial fifth roots of unity:
$$
1+\frac1z=e^{2\pi i k/5},\ \ k=1,2,3,4.
$$
So we get four solutions, namely
$$
z=\frac1{e^{2\pi i k/5}-1},\ \ k=1,2,3,4.
$$
</s>