I think this depends on what you're worried about, and what elements of the subject you think you might have trouble with. You mention that you thought real analysis was challenging, but you also ask specifically whether complex analysis is more rigorous than real analysis. So there are two possible aspects of the course one can look at: the level of mathematical rigor, which can differ depending on the style of the professor and/or textbook, and the difficulty of the mathematical concepts.
In terms of mathematical rigor, I would say it's rare that one would teach complex analysis more rigorously than one would teach real analysis, if it's being taught by the same professor (which, of course, is not given). That is not to say that complex analysis requires less rigor, but simply that because real analysis is in some sense more fundamental, in teaching it one often pays more attention to a lot of details.
In terms of mathematical concepts, I think it is typical for a student to say that complex analysis hearkens back to more familiar concepts learned in previous courses, e.g. basic calculus and geometry. In many respects, real analysis is a subject that has to deal with pathological issues: things like spaces that are path-connected but not connected, functions that are continuous everywhere but differentiable nowhere, metric spaces that are not complete, etc. This often makes it hard to visualize some of the concepts. Complex analysis, on the other hand, deals only with a nice space ($\mathbb{C} \cong \mathbb{R}^2$) and complex analytic functions, which turn out to be basically the nicest functions on the planet. You learn many geometric connections involving complex analytic functions, and often these connections can be visualized. I don't claim that complex analysis is an easier subject, but certainly the theory is much tighter and in many ways more elegant.
I hope this helps you in making your decision.
Denote the velocity components as
$$u = \frac{-2xy}{(x^2+y^2)^2}, \,\,\, v = \frac{x^2-y^2}{(x^2+y^2)^2}$$
The complex potential is an analytic function $\Psi(z) = \phi(x,y) + i \psi(x,y)$ of the complex variable $z = x+iy$ where the real part $\phi$ is the velocity potential and the imaginary part $\psi$ is the streamfunction.
Applying the Cauchy-Riemann equations we obtain
$$u = \frac{\partial \phi}{\partial x} = \frac{\partial \psi}{\partial y},\,\,\, v= \frac{\partial \phi}{\partial y} = -\frac{\partial \psi}{\partial x} $$
We can solve for $\phi$ by integrating first $u$ with respect to $x$,
$$\phi = \int \frac{-2xy}{(x^2+y^2)^2} \, dx = \frac{y}{x^2+y^2} + C(y)$$
Taking the derivative with respect to y we obtain
$$v = \frac{x^2 - y^2}{(x^2 + y^2)^2} = \frac{\partial \phi}{\partial y} = \frac{\partial}{\partial y}\frac{y}{x^2+y^2} + C'(y) = \frac{x^2 - y^2}{(x^2 + y^2)^2} + C'(y)$$
This implies that $C'(y) = 0$ and $C(y)$ is a constant which can be set arbitrarily to zero.
Thus, $\phi(x,y) = y/(x^2+y^2)$ and in a similar way we can find $\psi(x,y) = x/(x^2+y^2)$. The complex potential is then given by
$$\Psi(z) = \frac{y}{x^2+y^2} + i \frac{x}{x^2+ y^2}$$
We can compute the flux as a contour integral of the complex velocity $u -iv$, where the contour is the line segment with endpoints $1$ and $2i$. Equivalently, we can use a line integral in the real plane over the segment $C= \{(\mu(t),\lambda(t)) = (1-t, 2t): \, 0 \leqslant t \leqslant 1\}.$
Using the latter approach, we have
$$\text{flux} = \int_C \mathbf{v} \cdot \mathbf{n}\, ds,$$
where
$$\mathbf{n} = \left\langle\frac{-\lambda'(t)}{\sqrt{\mu'(t)^2 + \lambda'(t)^2}}, \frac{\mu'(t)}{\sqrt{\mu'(t)^2 + \lambda'(t)^2}} \right\rangle = \langle -2/\sqrt{3}, -1/\sqrt{3}\rangle,$$
$\mathbf{v} = \langle u(\mu(t), \lambda(t)), v(\mu(t), \lambda(t)) \rangle$ and $ds = \sqrt{\mu'(t)^2 + \lambda'(t)^2} dt = \sqrt{3} dt$.
You should be able to finish from here.
Best Answer
In the complex domain the theory of integration is quite different. In large part, this is due to Cauchy's integral formula. Long story short, there are many theorems in complex analysis which are not replicated in the real case. So, I would not agree these are the "same" thing.
That said, there is a fundamental theorem of calculus for both. As you say, the gradient vector field $\vec{F} = \nabla f$ has $\int_C \vec{F} \cdot d\vec{r} = f(B)-f(A)$ if the curve $C$ goes from $A$ to $B$ in a domain where the $f$ is defined near to $C$. Likewise, if $f(z) = g'(z)$ then $\int_C f(z) dz = g(z_1)-g(z_o)$ where $C$ goes from $z_o$ to $z_1$ again supposing $g$ is defined near $C$.
You mention a vector field from $\mathbb{C}^2$ to $\mathbb{C}^2$. However, the integral you wrote I think is just for functions from $\mathbb{C}$ to $\mathbb{C}$. These correspond to certain integrals of vector fields on $\mathbb{R}^2$. And, there are applications to fluid-flow or two-dimensional electrostatics. That said, I suspect the most natural justification for $\int_C f(z) dz$ is simply that it is the natural generalization of the Riemann integral if we replace real numbers and functions with their complex analogs.