Solve complex equation $z^3 = i$ I have this $z^3 = i$ complex equation to solve.

I begin with rewriting the complex equation to $a+bi$ format.

1 $z^3 = i = 0 + i$

2 Calculate the distance $r = \sqrt{0^2 + 1^2} = 1$

3 The angle is $\cos \frac{0}{1}$ and $\sin \frac{1}{1}$, that equals to $\frac {\pi}{2}$.

4 The complex equation can now be rewriten $w^3=r^3(cos3v+i\sin3v)$, $w^3 = 1^3(\cos \frac {\pi}{2} 3 +i \sin \frac {\pi}{2} 3)$ or $w^3 = e^{i \frac {\pi}{2} 3}$.

5 Calculate the angle $3 \theta = \frac {\pi}{2} + 2 \pi k$ where $k = 0, 1, 2$

6 $k = 0$,  $3 \theta = \frac {\pi}{2} + 2 \pi 0 = \frac {\pi}{6}$

7 $k = 1$,  $3 \theta = \frac {\pi}{2} + 2 \pi 1 = \frac {\pi}{6} + \frac {2 \pi}{3} = \frac {5 \pi}{6}$

8 $k = 2$,  $3 \theta = \frac {\pi}{2} + 2 \pi 2 = \frac {\pi}{6} + \frac {4 \pi}{3} = \frac {9 \pi}{6}$

So the angles are $\frac {\pi}{6}, \frac {3 \pi}{6},  \frac {9 \pi}{6}$ but that is no the correct answer. The angle of the complex equation should be $-\frac {\pi}{2}$ where I calculated it to $\frac {\pi}{2}$. I'm I wrong or is there a mistake in the book I'm using?

Thanks!
Way easier way; 

$$z^3=i \\
\iff z^3-i=0 \\
\stackrel{-i=i^3}{\iff}z^3+i^3=0 \\
\iff (z+i)(z^2-iz-1) = 0 \\
\iff z_1=-i,\;  z_2=\frac12 (i-\sqrt 3), \; z_3=\frac12 (i+\sqrt 3)$$

Disregard this answer if your exercises restrict you to trigonometric/polar form.
Use polar coordinates. 

$z^{3}=i=e^{i(\frac{\pi}{2}+2k\pi)}$, $k\in \mathbb{Z}$

And from here it is much simpler

EDIT: what I mean by much simpler...

The OP kind of used polar form, but not really. You should stock with polar form until the very end. The OP introduces $\theta$, $cosinus$ and $sinus$ functions etc. You mix everything up and forget that $(\frac{\pi}{2}+2\pi)$, divided by $3$, gives $\frac{5\pi}{6}$...