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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...

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    $\begingroup$ You are right about the real part of $z$ being $-1/2$. But for the argument, you need $5\arg(z+1)=5\arg(z)+2k\pi$ $\endgroup$
    – Empy2
    Nov 23, 2013 at 4:38
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    $\begingroup$ Logs are good for campfires. For complex numbers, not so much. I would note that $z\ne 0$, so we are solving $w^5=1$ where $w=\frac{z+1}{z}$. $\endgroup$
    – André Nicolas
    Nov 23, 2013 at 4:49
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    $\begingroup$ In the complex numbers, natural logarithm is not a function but rather multi-valued. Keeping track of all branches of the logarithm (they are like sheets that spiral about the singularity at the origin) is too much work when the basic problem to solve is a polynomial equation. $\endgroup$
    – hardmath
    Nov 23, 2013 at 5:13

5 Answers 5

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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?

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    $\begingroup$ Neat trick! I'll keep this one in mind in the future! $+1$ $\endgroup$
    – Cameron Williams
    Nov 23, 2013 at 4:38
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    $\begingroup$ Another way to look at this is to not notice the trick initially, leaving $5 z^4+10 z^3+10 z^2+5 z+1$, and then convert to the depressed quartic and notice that the $u$ term disappeared, too. $\endgroup$
    – Mark S.
    Nov 24, 2013 at 7:50
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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
$$ z=\frac1{e^{2\pi i k/5}-1},\ \ k=1,2,3,4. $$

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

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  • $\begingroup$ Why follows $$ \pars{1 + {1 \over z_{n}}}^{5} = \expo{2n\pi\ic}\,,\qquad n = 1, 2, 3, 4 $$ from $$ \pars{1 + {1 \over z}}^{5} = 1 $$ $\endgroup$
    – miracle173
    Nov 23, 2013 at 18:40
  • $\begingroup$ Note that $e^{i(2n\pi)} = \cos(2n\pi) + i \sin (2n \pi)$, but $\cos (2n \pi) = 1$ for all $n \in \mathbb{Z}$ and $\sin (2n \pi) = 0$. $\endgroup$
    – Mark Fantini
    Nov 24, 2013 at 0:24
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    $\begingroup$ Read in the voice of Captain Picard: "There are four roots." $\endgroup$
    – Doug McClean
    Nov 24, 2013 at 2:45
  • $\begingroup$ @miracle173: because $e^{2n\pi i}=1$ for any integer $n$. $\endgroup$
    – Martin Argerami
    Nov 24, 2013 at 16:50
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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

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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.

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    $\begingroup$ The logarithm is 1-1; the issue at hand here is that it is not a "function" in the more common meaning of the term, in that it is multivalued. $\endgroup$
    – Dustan Levenstein
    Nov 23, 2013 at 4:43
  • $\begingroup$ Cause: In complex numbers the map $\exp : z \mapsto e^z$ is not injective ("one-to-one"). But it is well-defined and single-valued. $\endgroup$
    – Jeppe Stig Nielsen
    Nov 24, 2013 at 12:42

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