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.
 A: $$z=(-16i)^{5/4}=32(-i)^{5/4}$$
Use 
$$-i=cis\left(\frac{3\pi}{2}+2k\pi\right)$$
Then 
$$(-i)^{5/4}=(-i)^{1/4}=cis\left(\frac{3\pi}{8}+\frac{1}{2}k\pi\right)$$
choose $k \in \{0,1,2,3\}$. So the solutions will be:
$$k=0 \rightarrow z_0=32\cdot cis\left(\frac{3\pi}{8}\right)\\
k=1 \rightarrow z_1=32\cdot cis\left(\frac{7\pi}{8}\right)\\
k=2 \rightarrow z_2=32\cdot cis\left(\frac{11\pi}{8}\right)\\
k=3 \rightarrow z_3=32\cdot cis\left(\frac{15\pi}{8}\right)
$$ 
A: 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.
A: First write $-16i$ in general polar form: $16e^{3\pi i/2+2\pi ik}$. Then $$(16e^{3\pi i/2+2\pi ik})^{5/4}=32e^{15\pi i/8+5\pi ik/2}=32e^{15\pi i/8+20\pi ik/8}.$$
By playing with values of $k$, we obtain the separate roots $32e^{\frac{3+4k}{8} \pi i}$.
Now, the discrepancy here is which of these is actually the principal root. I believe this is a matter of definitions. According to Wikipedia, the principal root is the one with the smallest positive argument. By that definition the answer has argument $3\pi/8$. However, I believe that Wolfram (and your calculator) has calculated the principal root by beginning with the smallest absolute value argument, which gives $16e^{-\pi i/2}$, and raising to the $5/4$, giving $32e^{-5\pi i /8}=32e^{11\pi i/8}$, which I believe is the answer you are looking for.
