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...
 A: How about, let $ u = z-1/2 $. Then in terms of $ u $ you have:
$$
\left(u+\frac{1}{2}\right)^5 = \left(u - \frac{1}{2}\right)^5
$$
Upon expansion:
$$
5 u^4 + \frac{5}{2} u^2 + \frac{1}{16} = 0
$$
A quadratic equation in $ u^2 $. Can you get it from here?
A: 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
$$
z=\frac1{e^{2\pi i k/5}-1},\ \ k=1,2,3,4.
$$
A: maybe also worth noting that the equation as given implies $|z+1|^2=|z|^2$ 

i.e. $(z+1)(z^*+1) = zz^*$ giving $z+z^*+1 = 0$

hence $\mathfrak{Re}(z)=-\frac12$,
which motivates user110781's neat substitution
A: In the complex numbers, the natural log is not one to one.  Because $e^{2\pi i}=1$ you can add $2\pi i$ to any log and get another one.  It is like the $\pm$ that shows up in the reals when you take a square root.  But you can view this as a quartic equation (the fifth powers cancel) which Alpha finds four roots for: $\frac 1{10}\left(-5\pm i\sqrt{5(5\pm 2\sqrt 5)}\right)$ where the plus or minus signs are all four choices.
A: $\newcommand{\+}{^{\dagger}}%
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 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
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 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
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$$
\pars{1 + {1 \over z_{n}}}^{5} = \expo{2n\pi\ic}\,,\qquad n = 1, 2, 3, 4
$$
$1 + 1/z_{n} = \expo{2n\pi\ic/5}\quad\imp\quad 1/z_{n} = \expo{2n\pi\ic/5} - 1$
$\quad\imp\quad z_{n} = \pars{\expo{2n\pi\ic/5} - 1}^{-1}$.
$$\color{#0000ff}{\large%
\mbox{There are four roots:}\quad
\left\lbrace%
\begin{array}{rcl}
z_{1} & = & {1 \over \expo{2\pi\ic/5} - 1}
\\
z_{2} & = & {1 \over \expo{4\pi\ic/5} - 1}
\\
z_{3} & = & {1 \over \expo{6\pi\ic/5} - 1}
\\
z_{4} & = & {1 \over \expo{8\pi\ic/5} - 1}
\end{array}\right.}
$$
