I can show that $(-1/2 + i\sqrt{3}/2)^3 = 1$. But why does this tell us that $(-1/2 - i\sqrt{3}/2)^3 = 1$? We have only covered what a complex number is, and what a conjugate is in my math class, and we haven't gone beyond that, so no more advanced tools please.
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Geometrically, what is the effect of complex multiplication, especially if the norm is one? Where are your two points and what happens if you cube them (multiply thrice)? – Integrand Oct 12 '20 at 23:12
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3Take conjugate on both sides and use the fact that conjugate of a product is the product of the conjugates. – Kavi Rama Murthy Oct 12 '20 at 23:12
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Algebraically, your first complex number $z$ is a root of $x^3-1$, which is a polynomial with real coefficients. It is easy to verify that if $p(x)$ is a polynomial with real coefficients, and $z\in\mathbb{C}$ is a root of $p(x)$, then so is $\overline{z}$, the conjugate of $z$ (evaluate at $z$, then take the conjugate of your whole expression). Your second complex number is nothing other than $(\overline{z})^3$.
Arturo Magidin
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If $z\in\mathbb{C}$ is a root, then $\overline{z}$ is also a root, because $\overline{z}^3=\overline{z.z.z}=\overline{z^3}=\overline{1}=1$.
Thus $(-1/2 + i\sqrt{3}/2)^3 = 1$ implies $\overline{(-1/2 + i\sqrt{3}/2)^3}=\overline{(-1/2 + i\sqrt{3}/2)}^3 =(-1/2 - i\sqrt{3}/2)^3 = \overline{1}=1$
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Small suggestion: a root of what? (A polynomial with real coefficients.) – Integrand Oct 12 '20 at 23:46
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$$(a+b)^3(a-b)^3=((a+b)(a-b))^3=(a^2-b^2)^3$$ Now take $a=1/2$ and $b=i\sqrt{3}/2$ to obtain $$(1)(1/2-i\sqrt{3}/2)^3 = (1/2+i\sqrt{3}/2)^3(1/2-i\sqrt{3}/2)^3=(1/4+3/4)^3=1$$
RobPratt
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