# When should I add $2k\pi$ to polar complex numbers

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?

• All you have shown is that the polar form of a complex number is not uniquely defined. It is true that $\frac{\sqrt 3}{2}+\frac{1}{2}i=\operatorname{cis}(\pi/6)$. It is also true that $\frac{\sqrt 3}{2}+\frac{1}{2}i=\operatorname{cis}(\pi/6+2k\pi)$ for all $k\in\Bbb Z$. Both of your answers are correct.
– Joe
Sep 6, 2021 at 11:43

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 $$cis(\pi/6+2k\pi), k\in\mathbb{Z}$$.

• So would $(\sqrt{3}/2 + 1/2i)^{16}=cis(8\pi/3+32k\pi)$? Sep 6, 2021 at 11:48
• @Relaxisys I would say $\left(\frac{\sqrt{3}}2 + \frac{1}{2}i\right)^{16} = -\frac12+\frac{\sqrt{3}}2i$ which is not quite equivalent to what you wrote. It $\left(\frac{\sqrt{3}}2 + \frac{1}{2}i\right)^{1/16}$ which is the more interesting multivalued case Sep 6, 2021 at 12:07
• @Henry so when should I change $2k\pi$ (if at all) when applying De Moivre's theorum? Sep 6, 2021 at 12:09

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.