Answer
There are 64 complex roots.
Two roots are real:
$cos~0^{\circ}+i~sin~0^{\circ} = 1$
$cos~180^{\circ}+i~sin~180^{\circ} = -1$
62 complex roots are not real.
Work Step by Step
$z = 1 = cos~0^{\circ}+i~sin~0^{\circ}$
$r = 1$ and $\theta = 0^{\circ}$
We can use this equation to find the 64th roots:
$z^{1/n} = r^{1/n}~[cos(\frac{\theta}{n}+\frac{360^{\circ}~k}{n})+i~sin(\frac{\theta}{n}+\frac{360^{\circ}~k}{n})]$, where $k \in \{0, 1, 2,...,n-1\}$
Note that there are 64 complex roots.
These two roots are real:
$cos~0^{\circ}+i~sin~0^{\circ} = 1$
$cos~180^{\circ}+i~sin~180^{\circ} = -1$
All other 62 complex roots have a non-zero imaginary part. Thus, 62 complex roots are not real.
