Complex Roots with improper fraction I'm having trouble with the following:
$(-16i)^{5/4}$
My calculations for the Principal root is:
$32(\cos (3\pi/2) * 5/4) + i \sin (3\pi/2)* 5/4))$
$=32(Cis (15\pi/8))$
This answer does not agree with the online calculators.  It gives a positive real value and the online calculators show a principal angle in Quadrant 3.
Confused on what happened here.

Let $z=-16i$ and $n=\dfrac54$. For solving you have to compute $r=|z|$ and argument $\theta$ where $\tan\theta=\dfrac{y}{x}$.
then
$r=|z|=|-16i|=16$ and argument $\theta=\dfrac{3\pi}{2}$. Then write
$$z_k=r^n(\cos n\theta+i\sin n\theta)$$
But argument adds with $2k\pi$ so we have
$$z_k=r^n\Big(\cos n(\theta+2k\pi)+i\sin n(\theta+2k\pi)\Big)$$
so
$$z_k=16^\dfrac54\Big(\cos\dfrac54(\dfrac{3\pi}{2}+2k\pi)+i\sin\dfrac54(\dfrac{3\pi}{2}+2k\pi)\Big)$$
and for $k=0,1,2,3$ write
$$z_k=32\Big(\cos(\dfrac{15\pi}{8}+\frac{5k\pi}{2})+i\sin(\dfrac{15\pi}{8}+\frac{5k\pi}{2})\Big)$$
finally for $k=0,1,2,3$ we conclude that
\begin{eqnarray}
k=0 &\Rightarrow& z_0=32\Big(\cos\dfrac{15\pi}{8}+i\sin\dfrac{15\pi}{8}\Big)=19.28+25.5i\\
k=1 &\Rightarrow& z_1=32\Big(\cos\dfrac{35\pi}{8}+i\sin\dfrac{35\pi}{8}\Big)=12.24+29.5i\\
k=2 &\Rightarrow& z_2=32\Big(\cos\dfrac{55\pi}{8}+i\sin\dfrac{55\pi}{8}\Big)=-29.5+12.24i\\
k=3 &\Rightarrow& z_3=32\Big(\cos\dfrac{75\pi}{8}+i\sin\dfrac{75\pi}{8}\Big)=-12.24-29.5i
\end{eqnarray}
This was step by step solving.
