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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 '13 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 '13 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 '13 at 5:13
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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 '13 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 '13 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 '13 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 '13 at 0:24
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    $\begingroup$ Read in the voice of Captain Picard: "There are four roots." $\endgroup$ – Doug McClean Nov 24 '13 at 2:45
  • $\begingroup$ @miracle173: because $e^{2n\pi i}=1$ for any integer $n$. $\endgroup$ – Martin Argerami Nov 24 '13 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 '13 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 '13 at 12:42

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