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Given the following equation:

$ \left(\tfrac{z-1}{z+1} \right)^3=i-1 $

It's possible to write:

$ \tfrac{z-1}{z+1}=\sqrt[3]{i-1} $

But the following step is corrent or I have to calculate the roots of $ i-1 $ first?

$ z-1=(z+1)\sqrt[3]{i-1} $

Stéphane Jaouen
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tizzzato
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  • What does $\sqrt[3]{i-1}$ mean? You can write whatever you want in mathematics provided you have defined it. So define it first. Is not it ? – Stéphane Jaouen Nov 03 '22 at 18:29
  • The three roots of the complex number. We calculate it using De Moivre's theorem – tizzzato Nov 03 '22 at 18:31
  • Let $z\in \Bbb C\setminus {-1}$. $z$ is a given number. So is $z-1$. So is $(\frac{z-1}{z+1})^3$. Can it be three things at once? – Stéphane Jaouen Nov 03 '22 at 18:35
  • Yes, \sqrt[3]{i-1} "contains" the three complex number that elevated to the power of three are equal to \sqrt[3]{i-1} – tizzzato Nov 03 '22 at 18:50

1 Answers1

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Let $S=\{z\in \mathbb C\setminus \{-1\}: (\frac{z-1}{z+1})^3=1-i\}$

Suppose $S$ is non-empty. Let then $z$ be in $S$. And let's set $Z=\frac{z-1}{z+1}$.

Then $Z^3=Z_0^3$, with $Z_0={2^{\frac16}}e^{\frac{i\pi}{4}}$.

Then $(\frac{Z}{Z_0})^3=1$ and $\frac{Z}{Z_0}\in \{1,j,j^2\}$.

Then necessarily, $Z\in \{Z_0,jZ_0,j^2Z_0\}$.

It then remains to solve the corresponding equations in $z$ and to ensure at the end that they agree and we'll get what we want, which is $S$.

Stéphane Jaouen
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  • Thanks! But can I write $z−1=(z+1)\sqrt[3]{i-1}$, $z(1-\sqrt[3]{i-1})=1+\sqrt[3]{i-1}$, $z=\frac{1+\sqrt[3]{i-1}}{1-\sqrt[3]{i-1}}$ and finding the roots only at this point? – tizzzato Nov 03 '22 at 19:39
  • as long as you haven't defined $\sqrt[3]{i-1}$, no : because here, as I showed it in my answer, there is no mathematical problem therefore useless to manipulate objects not clearly defined: it is not high mathematics that we do where we could allow some risk taking; but down-to-earth mathematics. – Stéphane Jaouen Nov 03 '22 at 19:48