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I need to convert the following complex number into polar form:
$ \sqrt{3}/2 + 1/2i$
After converting it into polar form, I got this:
$cis(\pi/6)$
Do I add $2k\pi$ like this: $ cis(\pi/6 +2k\pi)$ for my answer to be correct? Do I always need to add $2k\pi$? If not, when should I add it?
It depends on what is wanted. If you wat a polar form, then $cis(\pi/6)$ would be a correct answer. So are $cis(13\pi/6)$ and $cis(-11\pi/6)$. If instead, all polar forms must be rendered, then you would put $cis(\pi/6+2k\pi), k\in\mathbb{Z}$.
It kind of depends on your definition. $(\log,\,\mathrm{cis})$ is supposed to be the inversion of the group homomorphism $\mathbb C\to\mathbb C^\ast$ given by $z\to\exp(z)$.
But this map is not injective, it holds the same value if the imaginary part differs by a multiple of $2\pi$.
So now we can have two definitions: We might restrict the imaginary part of $z$ onto a smaller set, like $\mathbb R + \mathrm i[0,2\pi)$. Then we’d assume that $\mathrm {cis}(x) \in[0,2\pi)$.
The other idea is: If we factor $\mathbb C$ by the kernel of the map we get an isomorphism. The kernel of this are just $\mathrm i2\pi\mathbb Z$. Thus we’d have $\mathrm {cis}(x) \in \mathbb R / (2\pi\mathbb Z)$.
So in the first case you’d need to add some $2k\pi$ so that $0\leq \pi/6 < 2\pi$ (i.e. $k=0$). In the second case we’d just get the equivalence class $$ [\pi/6] = \{\pi/6+2k\pi;k\in\mathbb Z\}$$
To be clear: $\text{cis } x$ is shorthand for $\cos x+i\sin x,$ which is a single-valued function, which equals $e^{ix}$ when the latter too is being a single-valued function.
As such, no, you never need to write $\;\text{cis}\left(\frac\pi6 +2k\pi\right)\;$ in lieu of plain old $\;\text{cis}\left(\frac\pi6\right),\;$ the same way it is never necessary to replace $\;\sin\left(\frac\pi6\right)\;$ with $\;\sin\left(\frac\pi6+2k\pi\right).$
The issue of having to add the $\:(+2k\pi)\:$ typically arises when the multi-valued definition of $e^z$ is implicitly being invoked, for example, when obtaining roots of unity, and in this law:
for $\theta\in\mathbb R$ and $z\in\mathbb C,$ $$\left(e^{i\theta}\right)^z=e^{z\,i(\theta+2k\pi)}.$$
OP: so when should I change $2kπ$ (if at all) when applying De Moivre's Theorem?
De Moivre's Theorem applies only for integer powers, so doesn't invoke the multi-valued definition of $e^z,$ so you don't need to add $2k\pi$ when using applying it. Strictly speaking, adding $2k\pi$ is when taking the $n$th root, as pointed out in the preceding section.
