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.
$\require{cancel}$
The area element
On a sphere, using spherical coordinates, the area element is radial, moreover
$$
{\rm d}{\bf a} = r^2\sin\theta~{\rm d}\theta~{\rm d}\phi~\hat{\bf r}
$$
where $0 < \phi < 2\pi$ and $0 < \theta < \pi$
The argument of the integral
with this in mind
\begin{eqnarray}
{\bf B}\cdot {\rm d}{\bf a} &=& \left(\frac{\mu_0 m}{2\pi r^3}\cos\theta~\hat{\bf r} + \frac{\mu_0 m}{4\pi r^3}\sin\theta~\hat{\bf \theta} \right) \cdot \left(r^2\sin\theta~{\rm d}\theta~{\rm d}\phi~\hat{\bf r} \right) \\
&=& \frac{\mu_0 m}{2\pi r}\sin\theta\cos\theta{\rm d}\theta~{\rm d}\phi \cancelto{1}{(\hat{\bf r} \cdot \hat{\bf r})} + \frac{\mu_0 m}{4\pi r}\sin^2\theta {\rm d}\theta~{\rm d}\phi \cancelto{0}{(\hat{\bf r}\cdot \hat {\bf \theta})} \\
&=& \frac{\mu_0 m}{2\pi r}\sin\theta\cos\theta{\rm d}\theta~{\rm d}\phi
\end{eqnarray}
Everything together
\begin{eqnarray}
\oint_A {\rm d}{\bf a} \cdot {\bf B} &=& \int_{0}^{2\pi} {\rm d\phi} \int_{0}^{\pi}{\rm d}\theta~ \frac{\mu_0 m}{2\pi r}\sin\theta\cos\theta \\
&=& \left(\int_{0}^{2\pi} {\rm d}\phi\right)\left(\int_{0}^{\pi}{\rm d}\theta~ \frac{\mu_0 m}{2\pi r}\sin\theta\cos\theta \right) \\
&=& 2\pi\frac{\mu_0 m}{2\pi r} \int_0^\pi {\rm d}\theta~\sin\theta\cos\theta \\
&=& \frac{\mu_0 m}{r} \int_0^\pi {\rm d}\theta~\sin\theta\cos\theta \\
&=& 0
\end{eqnarray}
Best Answer
As per Biot-Savart's law, magnetic field due to a small element is: $$dB=\frac{\mu_0}{4\pi}\frac{i\vec{dl}\times \hat{r}}{r^2}$$ Notice that in polar coordinates, $\vec{dl}=dr\hat{r}+r\,d\theta \hat{\theta}$ i.e $$\vec{dl}\times \hat{r}=-r\,d\theta \hat{z}$$ where $\hat{z}$ is the direction pointing out of the plane. $$\Rightarrow \vec{B}=\frac{\mu_0 i}{4\pi} \int_{-\pi/2}^{\pi/2} \frac{-r\,d\theta}{r^2}\,\hat{z}$$ $$\begin{aligned} \Rightarrow \left|\vec{B}\right| &=\frac{\mu_0 i}{4\pi} \int_{-\pi/2}^{\pi/2} \frac{d\theta}{r}=\frac{\mu_0 i}{4a\pi} \int_{-\pi/2}^{\pi/2} \cos\theta(1+\sin^2\theta)\,d\theta\\ &=\frac{\mu_0 i}{4a\pi}\int_{-1}^1 (1+t^2)\,dt\,\,\,\,(\sin\theta=t) \\ &=\boxed{\dfrac{2}{3}\dfrac{\mu_0 i}{a\pi}} \\ \end{aligned}$$