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.

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

