<s>[INST] Find all the solutions in the complex field of the following equation: 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.
 [/INST] Ignoring the given hint, you can solve it by expanding out terms.
Take the square root of both sides (we get two options):
$$(1-z)^6=(1+z)^6\Longleftrightarrow (1-z)^3=\pm(1+z)^3$$


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


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*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:


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*$$\color{red}{z=0\space\space\space\vee\space\space\space z=\pm i\sqrt{3}\space\space\space\vee\space\space\space z=\pm\frac{i}{\sqrt{3}}}$$

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