Double integral (averaging) of an error function in a complex plane in an area such as circle

Question

I am going to first explain my goal in the introduction part and I will explain my difficulty in the problem statement:

Introduction:

I have my output $b_{out}(\Gamma_{out},\Gamma_1,\Gamma_2)$ that depends on two real variables $-1\le\Gamma_1\le1$ and $-1\le\Gamma_2\le1$, that I can choose, but $\Gamma_{out}$ is a complex variable that should ideally be zero but I cannot choose this variable. However, let's assume that each time this variable has this property: $|\Gamma_{out}|\le\Gamma_{out,max}\leq1$. For simplicity, let's assume $\Gamma_{out,max}=0.5$. Since I want to do my calculation in polar coordination, I write $\Gamma_{out}$ as

$$\Gamma_{out}=r \,{\mathrm{e}}^{\theta \,\mathrm{i}}$$

The relationship is as follows:

$$b_{out}(r \,{\mathrm{e}}^{\theta \,\mathrm{i}},\Gamma_1,\Gamma_2)=\frac{\sqrt{2}\,{\left(\Gamma_1 \,\Gamma_2 +1\right)}}{\Gamma_1 \,\Gamma_2 \,\mathrm{i}+\Gamma_2 \,r \,{\mathrm{e}}^{\theta \,\mathrm{i}} \,\mathrm{i}+2\,\mathrm{i}}$$

b_out_solved =

(2^(1/2)*(Gamma_1*Gamma_2 + 1))/(Gamma_1*Gamma_2*1i + Gamma_2*r_Gamma_out*exp(theta_Gamma_out*1i)*1i + 2i)

My goal is to find a closed-form form like $f(\Gamma_1,\Gamma_2)$ to have a relations ship between a variation of $b_{out}$ to $\Gamma_1$ and $\Gamma_2$ with the knowledge that $\Gamma_{out}$ can each time be a complex value within circle 0.5 in complex plane.

Therefore, I thought I should make a normalized error for each $\Gamma_1$ and $\Gamma_2$, then average over the area (circle) in $\Gamma_{out}$ plane. I write down my thought as follows. The ideal for $\Gamma_{out}$ is to be zero. The normalized error can be written as

$$err(r\,\mathrm{e}^{\theta \,\mathrm{i}},\Gamma_1,\Gamma_2)=\frac{|b_{out}(r \,{\mathrm{e}}^{\theta \,\mathrm{i}},\Gamma_1,\Gamma_2)-b_{out}(0,\Gamma_1,\Gamma_2)|^2}{|b_{out}(0,\Gamma_1,\Gamma_2)|^2}$$

Now if I average/ double integrate on the desired area of $\Gamma_{out}$, a complex plane, then I have the average error

$${\large{} \left.f(\Gamma_{1},\Gamma_{2})\right|_{|\Gamma_{out}|\leq0.5}=\frac{1}{\pi(0.5)^{2}}\int_{0}^{\Gamma_{out,max}=0.5}\int_{0}^{2\pi}err(r\,\mathrm{e}^{\theta\,\mathrm{i}},\Gamma_{1},\Gamma_{2})\,r\mathrm{d}\theta\mathrm{d}r }{\large} $$

Questions:

1. My first question is Does it make sense? I hope what I did makes sense for you. Please let me know if it does not.

2. In the normal non-complex plane, we can simply use $r\mathrm{d}\theta\mathrm{d}r$. I have a complex plane here, can I still use $r\mathrm{d}\theta\mathrm{d}r$?

3. I tried to implement what I explained, step by step, in MATLAB symbolic environment. My code was:

b_out_ideal = subs(b_out_solved,[r_Gamma_out,theta_Gamma_out],[0,0])
eqn_fom_1 = abs(b_out_solved-b_out_ideal)^2/abs(b_out_ideal)^2
eqn_fom_2 = int(eqn_fom_1*r_Gamma_out,theta_Gamma_out,0,2*pi)
eqn_fom_3 = int(eqn_fom_2,r_Gamma_out,0,0.5)
eqn_fom_4 = sqrt(1/(pi*0.5^2)*eqn_fom_3)
fsurf((eqn_fom_3),[-1 1 -1 1],'MeshDensity',10)

Unfortunately, fsurf is very slow and it has some division by zero.

There are some scenarios that this will happen like considering $\Gamma_2$=0 in the following results from MATLAB:

eqn_fom_3 =

$$ \begin{array}{l} \left\lbrace \begin{array}{cl} \sqrt{\frac{5734161139222659\,\pi \,{\Gamma_2 }^2 \,\int_0^{\frac{1}{2}} \frac{{r_{\Gamma ,\textrm{out}} }^3 }{\sigma_9 +4\,\Gamma_1 \,\Gamma_2 -\sigma_7 +4}\textrm{d}r_{\Gamma ,\textrm{out}} }{2251799813685248}} & \;\textrm{if}\;\;\sigma_1 \in \left\lbrack 0,\frac{1}{2}\right\rbrack \\ \sqrt{-\frac{5734161139222659\,\pi \,{\Gamma_2 }^2 \,{\left(\sigma_2 -{\left(\lim_{r_{\Gamma ,\textrm{out}} \to {-\sigma_1 }^{_- } } \;\sigma_4 \right)}+{\left(\lim_{r_{\Gamma ,\textrm{out}} \to {-\sigma_1 }^{_+ } } \;\sigma_4 \right)}-\sigma_6 +\sigma_5 \right)}}{2251799813685248}} & \;\textrm{if}\;\;\sigma_1 \not\in \left\lbrack 0,\frac{1}{2}\right\rbrack \wedge \sigma_1 \in \sigma_3 \\ \sqrt{-\frac{5734161139222659\,\pi \,{\Gamma_2 }^2 \,{\left(\sigma_2 -\sigma_6 +\sigma_5 \right)}}{2251799813685248}} & \;\textrm{if}\;\;\sigma_1 \not\in \left\lbrack 0,\frac{1}{2}\right\rbrack \wedge \sigma_1 \not\in \sigma_3 \end{array}\right.\\ \mathrm{}\\ \textrm{where}\\ \mathrm{}\\ \;\;\sigma_1 =\frac{\Gamma_1 \,\Gamma_2 +2}{\Gamma_2 }\\ \mathrm{}\\ \;\;\sigma_2 =\frac{1}{8\,{\Gamma_2 }^2 }\\ \mathrm{}\\ \;\;\sigma_3 =\left\lbrack -\frac{1}{2},0\right\rbrack \\ \mathrm{}\\ \;\;\sigma_4 =-\frac{{r_{\Gamma ,\textrm{out}} }^2 }{2\,{\Gamma_2 }^2 }-\frac{\mathrm{log}\left(-\sigma_9 -4\,\Gamma_1 \,\Gamma_2 +\sigma_7 -4\right)\,\sigma_8 }{2\,{\Gamma_2 }^4 }\\ \mathrm{}\\ \;\;\sigma_5 =\frac{\mathrm{log}\left(-\sigma_9 -4\,\Gamma_1 \,\Gamma_2 +\frac{{\Gamma_2 }^2 }{4}-4\right)\,\sigma_8 }{2\,{\Gamma_2 }^4 }\\ \mathrm{}\\ \;\;\sigma_6 =\frac{\mathrm{log}\left(-\sigma_9 -4\,\Gamma_1 \,\Gamma_2 -4\right)\,\sigma_8 }{2\,{\Gamma_2 }^4 }\\ \mathrm{}\\ \;\;\sigma_7 ={\Gamma_2 }^2 \,{r_{\Gamma ,\textrm{out}} }^2 \\ \mathrm{}\\ \;\;\sigma_8 =\sigma_9 +4\,\Gamma_1 \,\Gamma_2 +4\\ \mathrm{}\\ \;\;\sigma_9 ={\Gamma_1 }^2 \,{\Gamma_2 }^2 \end{array}$$

4. Unfortunately, I do not know what to do for the next step. How should I solve this problem? Am I doing something wrong and how to solve it?

I would be grateful to have some feedback.