You can go ahead with your Cauchy-Riemann approach, but I also know of a geometric argument which I like a lot. Suppose $f(z)$ is nonconstant on $D$, and let $a \in D$ such that $f'(a) \neq 0$. By the definition of derivative, we know that $f(a + h) = f(a) + hf'(a) + o(h)$.
Choose $\alpha$ in $\Bbb{C}$ so that when $\alpha$ is considered as a vector in $\Bbb{R}^2$, $\alpha f'(a)$ is a unit vector perpendicular to $f(a)$. Let $h = r \alpha$, with $r$ real and positive. Note that $arg(f(a) + hf'(a)) \neq arg(f(a)).$ As $r$ goes to zero we see that $f(a + h)$ lies in a disk of radius $o(r)$ around $f(a) + h f'(a)$. For $r$ sufficiently small, the radius of this disc is smaller than the distance between $f(a) + h f'(a)$ and $f(a)$.
But $f(a)$ is the closest point to $f(a) + h f'(a)$ on the line $arg(z) = arg(f(a))$, since $h f'(a)$ is perpendicular to that line. It follows that when $h$ is sufficiently small, the disk and the line do not intersect. But $f(a + h)$ lies in the disk, and therefore cannot have the same argument as $f(a)$. This is a contradiction.
We conclude that if $arg(f(z))$ is constant, $f'(z)$ is identically zero on $D$. Assuming that $D$ is connected this implies that $f$ is a constant function there.
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
Observe that $-1+\sqrt{3}i=2e^{\frac{2\pi i}{3}}$. Then the roots of the equation are $$ z=\sqrt[4]{2}e^{\frac{\pi i}{6}+\frac{k\pi}{2}}, $$ where $k=0,1,2,3$. Thus $r=\sqrt[4]{2}$. Substitute $k$, then we get $z=\sqrt[4]{2}e^{\frac{\pi i}{6}}$, $z=\sqrt[4]{2}e^{\frac{2\pi i}{3}}$, $z=\sqrt[4]{2}e^{\frac{7\pi i}{6}}$, and $z=\sqrt[4]{2}e^{\frac{5\pi i}{3}}$. There is no root whose argument is $\frac{11\pi}{12}$. What happened? Because your attempt has a flaw: Since $r^4 \cos 4\theta=-1$ and $r^4\sin 4\theta=\sqrt{3}$, we get $\tan 4\theta=-\sqrt{3}$, but it doesn't imply that $4\theta = -\frac{\pi}{3}$. If it is true, then $r^4 \cos4\theta$ cannot be negative!