Is $Log(i^{1/3})=\frac{1}{3}Log(i)$? The definition of general complex log for any non-zero complex number $z$ is 
$$Log(z)=\log|z|+i[\arg(z)+2m\pi], m\in \mathbb{Z}$$
With this, if $n\in \mathbb{N}$ then 
$Log(z^{1/n})=\frac{1}{n} Log(z)$ holds for all non-zero complex number $z$. 
I verified this for $Log(i^{1/2})=\frac12 Log(i)$ successfully but could not make up with the following: $$Log(i^{1/3})=\frac13 Log(i)$$
Let me show what I have done and where I got stuck. 
Since $i=\cos(2n\pi+\frac{\pi}{2})+i \sin(2n\pi+\frac{\pi}{2}), n\in \mathbb{Z}$ then by De-Moivre' s theorem, we have 
\begin{align}
i^{1/3}= & \cos\left(\frac{2n\pi+\frac{\pi}{2}}{3}\right)+i\sin\left(\frac{2n\pi+\frac{\pi}{2}}{3}\right), n=0,1,2 \\
=& \cos\left(\frac{(4n+1)\pi}{6}\right)+i\sin\left(\frac{(4n+1)\pi}{6}\right), n=0,1,2\\       
= &\begin{cases}
            \cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6})    \\
            \cos(\frac{5\pi}{6})+i\sin(\frac{5\pi}{6})    \\
            \cos(\frac{9\pi}{6})+i\sin(\frac{9\pi}{6})   
           \end{cases}\\
= &\begin{cases}
            \cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6})    \\
            \cos(\pi-\frac{\pi}{6})+i\sin(\pi-\frac{5\pi}{6})    \\
            -\cos(\frac{\pi}{2})-i\sin(\frac{\pi}{2})   
           \end{cases}\\
= &\begin{cases}
            \frac{\sqrt{3}}{2}+i\frac{1}{2}    \\
            -\frac{\sqrt{3}}{2}+i\frac{1}{2}    \\
            0-i   
           \end{cases}
\end{align}
Now LHS:
\begin{align}
\frac13 Log(i)=&\frac13[\log|i|+i\{\arg(i)+2n_1 \pi\}], n_1\in \mathbb{Z}\\
              =&\frac13(2n_1\pi+\frac{\pi}{2}), n_1\in \mathbb{Z}\\
              =&(4n_1+1)\frac{\pi i}{6}, n_1\in \mathbb{Z}
\end{align}
whereas we see that 
\begin{align}
 &Log(i^{1/3})\\
=&\begin{cases}
  \log|\frac{\sqrt{3}}{2}+i\frac{1}{2}|+i[\arg(\frac{\sqrt{3}}{2}+i\frac{1}{2})+2m_1\pi]\\
  \log|-\frac{\sqrt{3}}{2}+i\frac{1}{2}|+i[\arg(-\frac{\sqrt{3}}{2}+i\frac{1}{2})+2m_2\pi]\\
  \log|-i|+i[\arg(-i)+2m_3\pi]
  \end{cases}\\
=&\begin{cases}
  i[\frac{\pi}{6}+2m_1\pi]\\
  i[\frac{5\pi}{6}+2m_2\pi]\\
  i[\frac{-\pi}{2}+2m_3\pi]
  \end{cases}, m_1, m_2, m_3\in \mathbb{Z}
\end{align}
And here I got stuck. I don't know how to finish. Any help will be appreciated. 
 A: You have 
$$ i = \exp\left(i\frac{\pi}{2} + i2n\pi\right) $$
Taking the third power yields
$$ i^{1/3} = \exp\left(i\frac{\pi}{6} + i \frac{2n\pi}{3} i\right) $$
Since $|i| = |i^{/3}| = 1$, taking the log yields
$$ \log (i^{1/3}) = i\left(\frac{\pi}{6} + \frac{2n\pi}{3} \right) $$
On the other hand, we have
$$ \frac{1}{3}\log (i) = \frac{i}{3} \left(\frac{\pi}{2} + 2n\pi \right) = i \left( \frac{\pi}{6} + \frac{2n\pi}{3} \right) $$
Does this help?
EDIT: Continuing from your work, you can show that the 3 arguments for $\log(i^{1/3})$ are evenly spaced by an angle of $2\pi/3$. Therefore they can all be combined as
$$ \left\{\begin{aligned} i\left(-\frac{\pi}{6} + 2m_1\pi\right) \\ i\left(\frac{5\pi}{6} + 2m_2\pi\right) \\ i\left(-\frac{\pi}{2} + 2m_3\pi\right) \end{aligned}\right. = i\left(\frac{\pi}{6} + \frac{2n\pi}{3}\right) = i(4n+1)\frac{\pi}{6} $$
where $n$ is mapped by every alternating triplet of $(m_1,m_2,m_3)$. For example $m_3 = 0, m_1 = 0, m_2 = 0 \to n = -1,0,1$, etc
A: Let $z^3-i=0$, then, by cube roots of unity, its roots are $-i, -i\omega, -i\omega^2$, for $\omega=e^{i2\pi/3}$ the cube root of unity. These values translate to $e^{i3\pi/2}, e^{i\pi/6}, e^{i5\pi/6}$.
Hint: I am showing this proof for $z=e^{i3\pi/2}$, but the procedure is same for the other two values.
So, $$\ln i^{1/3}=i3\pi/2$$
So, if $1/3\cdot \ln i=i3\pi/2$
Then, $\ln i=i9\pi/2$ that is  $i=e^{i9\pi/2}=e^{i\pi/2+i4\pi}=e^{i\pi/2}$ which is true. Hence, proved.
Where you went wrong? You were using DeMoivre's theorem and expanding to $\cos$ and $\sin$, when you should have used the Euler form instead. That is, you were using the right theorem in the wrong place.
Why should I use the Euler's form? Well, you already have $\ln$ in your question, and we can easily manipulate complex numbers using $\ln$, when they are in Euler form.
Hope it helps!
