Can you Simplify This Complex Expression? Let $a,b\in\mathbb{R}$ and $i=\sqrt{-1}$.  Does the expression $(a+i b)^{1/3} + (a-i b)^{1/3}$ simplify to a real valued expression defined solely in terms $a$ and $b$? 
 A: Let $a+bi = z$ and assume that $|z| = 1$. Then $z = \cos \theta + i \sin \theta$. Then the sum is equal to $\cos \theta/3+i \sin \theta/3+ \cos \theta/3-i \sin \theta/3 = 2 \cos \theta/3$ which is real. 
A: We can also look at this geometrically, knowing (from DeMoivre) how the roots of a complex number are arranged on the complex plane.  The number $ \ z = a + bi \ $ has its three cube roots arranged 120º apart on a circle with a radius given by the cube root of the modulus of $ \ z \ $  [marked in blue].  The complex conjugate $ \ \overline{z} \ $ has its three cube-roots [marked in red] also spaced 120º apart on this same circle (since it has the same modulus as $ \ z \ $ , its cube-roots must have the same moduli as the cube-roots of $ \ z \ $ ) .
Since $ \ \overline{z} \ $ has an argument which is the negative of that of $ \ z \ $ , its cube-roots will have arguments which are the negative of those for the cube-roots of $ \ z \ $ .  So the cube-roots of $ \ \overline{z} \ $ are the complex conjugates of those of $ \ z \ $ . The sum of each cube-root and its corresponding conjugate is then a real number.
This graph is not offered as a "proof-by-picture" , but simply to illustrate the situation.

