There is no such an angle except using the inverse trigonometri functions.
Otherwise, you can approximate it since it looks to be quite close to $\frac \pi 4$ and using a truncated Taylor expansion around this value, you would get
$$\tan^{-1}(x)=1+2 \left(x-\frac{\pi }{4}\right)+O\left(\left(x-\frac{\pi }{4}\right)^2\right)$$ and then, ignoring the higher order terms, you could solve
$$1+2 \left(x-\frac{\pi }{4}\right)=\frac{\sqrt{3}}2\implies x=\frac{\sqrt{3}+\pi-2 }{4}\approx 0.7184 $$ while the exact value would be $0.7137$.
You could also use very nice approximations for $\sin(x)$ or $\cos(x)$ (have alook here).
Edit
Sooner or later, you will learn that, better than with Taylor series, functions can be locally approximated using Padé approximants. Using the simplest around $x=a$, we have
$$\tan(x)=\frac{1+(x-a)}{1-(x-a)}$$ Using it for your problem, we just need to solve $$\frac{1+(x-\frac \pi 4)}{1-(x-\frac \pi 4)}=\frac{\sqrt{3}}2\implies x=\frac \pi 4+4 \sqrt{3}-7\approx 0.7136$$ which is much better.
Let $a,b\in\mathbb{R}$ so that $$\sqrt{i+1} = a+bi$$
$$ i+1 = a^2 -b^2 +2abi $$
Equating real and imaginary parts, we have
$$2ab = 1$$
$$a^2 -b^2 = 1$$
Now we solve for $(a,b)$.
$$
\begin{align*}
b &= \frac{1}{2a}\\\\
\implies \,\,\, a^2 - \left(\frac{1}{2a}\right)^2 &= 1 \\\\
a^2 &= 1 + \frac{1}{4a^2}\\\\
4a^4 &= 4a^2 + 1\\\\
4a^4 - 4a^2 -1 &= 0 \\\\
\end{align*}
$$
This is a quadratic in $a^2$ (it's also a quadratic in $2a^2$, if you prefer!), so we use the quadratic formula:
$$a^2 = \frac{4 \pm \sqrt{16-4(4)(-1)}}{2(4)}$$
$$a^2 = \frac{1 \pm \sqrt{2}}{2}$$
Here we note that $a$ is real, so $a^2>0$, and we discard the negative case:
$$a^2 = \frac{1 + \sqrt{2}}{2}$$
$$a = \pm \sqrt{\frac{1 + \sqrt{2}}{2}}$$
$$ b = \frac{1}{2a} = \pm \sqrt{\frac{\sqrt{2}-1}{2}}$$
This gives what you can call the principal root:
$$\sqrt{i+1} = \sqrt{\frac{1 + \sqrt{2}}{2}} + i\sqrt{\frac{\sqrt{2}-1}{2}} $$
As well as the negation of it:
$$-\sqrt{i+1} = -\sqrt{\frac{1 + \sqrt{2}}{2}} + i\left(-\sqrt{\frac{\sqrt{2}-1}{2}}\right) $$
Finally, substituting either of these into your expression $$z=2i \pm \sqrt{i+1}$$ will give you $\text{Re}(z)$ and $\text{Im}(z)$.
At that point, as you noted in your question, conversion to polar coordinates is straightforward.
Best Answer
Note that
thus
$$4\left(-\cos(\frac\pi3)-\sin(\frac\pi3)i\right)=-2-2\sqrt3\,i$$