Firstly my apologies for anything that should be in LaTex format correctly, I gave it a valiant effort. I have been asked to solve the equation: $(1-z)^6 = (1+z)^6$ A hint given states: do not multiply out!. If the equation was in a more suitable for I would have multiplied by $(1-z)$ and solve it using De Moivre's theorem to find the roots but I am simply unable to get to such a point. I thought about taking the square root of both sides in order to make it simpler.

  • 2
    $\begingroup$ Hint: You can find all the solutions to $w^6=1$. That gives all the solutions to your equation, by letting $w=\dots$. $\endgroup$ – David C. Ullrich May 13 '16 at 16:59

Ignoring the given hint, you can solve it by expanding out terms.

Take the square root of both sides (we get two options): $$(1-z)^6=(1+z)^6\Longleftrightarrow (1-z)^3=\pm(1+z)^3$$

  • For $(1-z)^3=(1+z)^3$:

$$(1-z)^3=(1+z)^3\Longleftrightarrow$$ $$(1-z)^3=z^3+3z^2+3z+1\Longleftrightarrow$$ $$(1-z)^3-(z^3+3z^2+3z+1)=0\Longleftrightarrow$$ $$-6z-2z^3=0\Longleftrightarrow$$ $$-2z(z^2+3)=0\Longleftrightarrow$$ $$z(z^2+3)=0$$

So, we get two options, $z=0$ or:

$$z^2+3=0\Longleftrightarrow z^2=-3\Longleftrightarrow z=\pm i\sqrt{3}$$

  • For $(1-z)^3=-(1+z)^3$:

$$(1-z)^3=-(1+z)^3\Longleftrightarrow$$ $$(1-z)^3=-z^3-3z^2-3z-1\Longleftrightarrow$$ $$(1-z)^3+z^3+3z^2+3z+1=0\Longleftrightarrow$$ $$6z^2+2=0\Longleftrightarrow6z^2=-2\Longleftrightarrow z^2=-\frac{1}{3}\Longleftrightarrow z=\pm\frac{i}{\sqrt{3}}$$

So, the solutions we've got are:

  • $$\color{red}{z=0\space\space\space\vee\space\space\space z=\pm i\sqrt{3}\space\space\space\vee\space\space\space z=\pm\frac{i}{\sqrt{3}}}$$
  • $\begingroup$ I haven't got an official solutions booklet but wolfram alpha seemed to confirm this, you are star. Merci beaucoup! $\endgroup$ – Pstar007 May 13 '16 at 17:15
  • $\begingroup$ @Pstar007 I'm glad that I could help. You're welcome (de rien :) ) $\endgroup$ – Jan May 13 '16 at 17:36

If you let $w=(1+z)/(1-z)$ (noting that $z=1$ is not a solution) your equation becomes $w^6=1$. The solutions to that are $$w=e^{2\pi ik/6},\quad k=0,\dots,5.$$So the solutions to your equation are the same as the solutions to the six linear equations $$1+z=e^{2\pi ik/6}(1-z)\quad(k=0,\dots,5).$$Except one of those equations has no solution.

  • $\begingroup$ Took my head slightly longer to understand this but worked it through and voila I got the same answers that Jan Eerland provided, cheers mate! $\endgroup$ – Pstar007 May 13 '16 at 17:33

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