I am trying to show that $\log(i^2) \ne 2\log(i)$ and that $\log(i^{1/2}) = 1/2\log(i)$ but I am having difficulties.
I know that $\log(z) = \ln|z| + i\arg(z)$ but when playing around with these equations, I keep getting the wrong answers. I know I must be doing something wrong but I am not sure where.
For the first one: $\log(i^2) = \log(-1) = \ln|-1| + i\arg(-1)$. Since $\arg(-1) = \pi + 2k\pi$ we get that $\log(i^2) = i(\pi + 2k\pi)$.
I also get that $2\log(i) = 2 (\ln|i| + i\arg(i)) = 2(i(2k\pi+\pi/2)) = i((4k+1))\pi$ which effectively gives me the same thing.
Similarly, for the second one, I ended up with $\log(i^{1/2}) = i(2k\pi + \pi/4)$ and $1/2\log(i) = 1/2[i(2k\pi + \pi/2)]$, which is not the same thing.
Where have I made a mistake?
You seem to have gotten the first one right ($2k +1 \neq 4k+1$ unless $k=0$), but can't conclude. Since a branch cut is defined as a right inverse, it may be easier to reason starting with exponentials.
$i^2 = e^{i\pi + 2k \pi i} $, and $i=e^{i\pi/2 + 2k\pi i}$, so the set of possible logarithms are $$ \log (i^2) = i(\pi + 2k \pi ) , \ \log i = i(\pi/2 + 2k\pi ) , \quad k\in\mathbb Z$$ therefore, with $\lambda A := \{ \lambda a : a \in A\}$, $$ \log (i^2) = i(\pi + 2k \pi ) , \ 2\log i = i(\pi + 4k\pi ) , \quad k\in\mathbb Z$$ and so $\log (i^2)$ and $2\log i$ describe different sets.
If we fix a branch cut of log along the negative imaginary axis that corresponds to $k=1$ in the above sets, the log becomes single valued, and we find that $$\log (i^2) = 3\pi i, \quad 2\log i = 5\pi i $$ but if you were to use one corresponding to $k=0$, you would see that $\log(i^2) = \pi i= 2\log i$.
For $i^{1/2}$, with the set valued definition $i^{1/2} := e^{1/2 \log i},$ we see that the set of possible values is $$i^{1/2} = e^{\frac12(i\pi/2 + 2k\pi i)} = e^{i\pi/4+k\pi i} $$
which agrees with the high-school knowledge that there are two square roots for every number. Thus, the set of all possible logarithms are $$ \log (i^{1/2}) = i\pi / 4 + k \pi i , \quad k\in\mathbb Z $$ which is equal as a set to $\frac12 \log i$, but it isn't equal to the set $\log (e^{i\pi/4}) = i\pi/4 + 2k\pi i \ (k\in \mathbb Z)$ or the set $\log (-e^{i\pi/4}) = i\pi/4 + (2k+1)\pi i \ (k\in \mathbb Z)$.