Solving $(z+1)^5 = z^5$

Question

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...

Answer 1

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?

Answer 2

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
$$ z=\frac1{e^{2\pi i k/5}-1},\ \ k=1,2,3,4. $$

Answer 3

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\dd}{{\rm d}}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\ic}{{\rm i}}% \newcommand{\imp}{\Longrightarrow}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert #1 \right\vert}% \newcommand{\yy}{\Longleftrightarrow}$ $$ \pars{1 + {1 \over z_{n}}}^{5} = \expo{2n\pi\ic}\,,\qquad n = 1, 2, 3, 4 $$ $1 + 1/z_{n} = \expo{2n\pi\ic/5}\quad\imp\quad 1/z_{n} = \expo{2n\pi\ic/5} - 1$ $\quad\imp\quad z_{n} = \pars{\expo{2n\pi\ic/5} - 1}^{-1}$. $$\color{#0000ff}{\large% \mbox{There are four roots:}\quad \left\lbrace% \begin{array}{rcl} z_{1} & = & {1 \over \expo{2\pi\ic/5} - 1} \\ z_{2} & = & {1 \over \expo{4\pi\ic/5} - 1} \\ z_{3} & = & {1 \over \expo{6\pi\ic/5} - 1} \\ z_{4} & = & {1 \over \expo{8\pi\ic/5} - 1} \end{array}\right.} $$

Answer 4

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

Answer 5

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.