Complex Numbers Exercise

Question

If $a,b,c$ are complex numbers with $a+b+c=0$ and $\|a\|=\|b\|=\|c\| = r>0$ then prove that $$a^{2^n} + b^{2^n} + c^{2^n} = 0$$

Any ideas?

Thanks!

Answer 1

We have $$ a + b + c = 0 $$ Taking the conjugate and multiplying by $abc$, $$ abc \overline{a} + abc \overline{b} + abc \overline{c} = 0 $$ i.e. $$ \newcommand{\abs}[1]{\left| #1 \right|} bc \abs{a}^2 + ac \abs{b}^2 + ab \abs{c}^2 $$ But then $\abs{a} = \abs{b} = \abs{c}$, so $$ ab + bc + ca = 0 $$ It follows that $$ a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) = 0 $$ In summary, we have shown that if $a + b + c = 0$ and $a, b, c$ have the same norm, $a^2 + b^2 + c^2 = 0$. Since $a^2, b^2, c^2$ again have the same norm, it follows by induction that $$ a^{2^n} + b^{2^n} + c^{2^n} = 0 $$ for all $n$.

Answer 2

Think of $a,b,c$ as vectors in the complex plane. Putting the tail of $a$ at zero, then putting the tail of $b$ to the tip of $a$, then putting the tail of $c$ to the tip of $b$, since $a+b+c=0$ it follows that the tip of $c$ is zero. Thus, the vectors form a triangle. This is an equilateral triangle of side length $r$. Now think about what happens geometrically when you raise a complex number to an integer power $k$. Then draw the triangle with side lengths $a^k,b^k,c^k$.