Complex Algebra/Examples/(1+z)^5 = (1-z)^5

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Example of Complex Algebra

The roots of the equation:

$\paren {1 + z}^5 = \paren {1 - z}^5$

are:

$z = \set {\dfrac {\omega^k - 1} {\omega^k + 1}: k = 0, 1, 2, 3, 4}$

That is:

$z = \set {0, \dfrac {\omega - 1} {\omega + 1}, \dfrac {\omega^2 - 1} {\omega^2 + 1}, \dfrac {\omega^3 - 1} {\omega^3 + 1} , \dfrac {\omega^4 - 1} {\omega^4 + 1} }$

where:

$\omega = \cis \dfrac {2 \pi i} 5$


Proof

We have:

\(\ds \paren {1 + z}^5\) \(=\) \(\ds \paren {1 - z}^5\)
\(\ds \leadsto \ \ \) \(\ds \frac {\paren {1 + z}^5} {\paren {1 - z}^5}\) \(=\) \(\ds 1\)
\(\ds \leadsto \ \ \) \(\ds \frac {1 + z} {1 - z}\) \(=\) \(\ds 1, \cis \dfrac {2 i \pi} 5, \cis \dfrac {4 i \pi} 5, \cis \dfrac {6 i \pi} 5, \cis \dfrac {8 i \pi} 5\)


Let $\cis \dfrac {2 \pi i} 5 = \omega$ as suggested.

Thus:

\(\ds \frac {1 + z} {1 - z}\) \(=\) \(\ds 1, \omega, \omega^2, \omega^3, \omega^4\)
\(\ds \) \(=\) \(\ds \omega^n: n = 0, 1, \dotsc, 4\)


So:

\(\ds 1 + z\) \(=\) \(\ds \paren {1 - z} \omega^n\)
\(\ds \leadsto \ \ \) \(\ds 1 + z\) \(=\) \(\ds \omega^n - \omega^n z\)
\(\ds \leadsto \ \ \) \(\ds z \paren {\omega^n + 1}\) \(=\) \(\ds \omega^n - 1\)
\(\ds \leadsto \ \ \) \(\ds z\) \(=\) \(\ds \frac {\omega^n - 1} {\omega^n + 1}: n = 0, 1, \ldots, 4\)


When $n = 0$:

$\omega^n = 1$

and thus:

$z = \paren {1 - 1} / \paren {1 + 1}$

So:

$z = 0, \dfrac {\omega - 1} {\omega + 1}, \dfrac {\omega^2 - 1} {\omega^2 + 1}, \dfrac {\omega^3 - 1} {\omega^3 + 1}, \dfrac {\omega^4 - 1} {\omega^4 + 1}$

$\blacksquare$


Sources

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