I am not exactly sure answering this question rather than the former is the best course of action, but this question is the more fleshed-out one and is more amenable to the answer I gave you in chat, which I am expanding upon here.

No, the transformation $z\mapsto \bar{w}_2z$ has nothing to do with degenerate triangles. In fact, if you start with a degenerate triangle (where the three endpoints are collinear), applying the map to the triangle will just yield *another* degenerate triangle - it doesn't change anything. This should be more apparent once you see what multiplication does to the complex plane. But it's not like degenerate triangles pose any issues in the first place!

The idea is that we want to use the formula Area = Base x Height / 2, however it isn't entirely obvious what the "height" of the triangle will be for any choice of edge as "base." This can be fixed, however, if we *rotate* the triangle (around the origin) so that one of its edges is on the real axis - in this case, the base is the length of that edge, and the height is the imaginary part of the third point.

Let's review some basic facts of the complex plane and its geometry. Every nonzero number $\alpha\in\bf C$ can be written in polar form, $\alpha=re^{i\theta}$. Here, $r=|\alpha|$ is the modulus, or magnitude - the distance between $\alpha$ and $0$. And $\theta$ is the angle made between the line segment from $0$ to $\alpha$ and the positive real axis. Note that $e^{i\theta}$ is the rescaling of $\alpha$ that is on the unit circle in the same direction.

So if $\alpha=re^{i\theta}$ and $\beta=se^{i\phi}$, then $\alpha\beta=rse^{i(\theta+\phi)}$. In words, this means that the effect of multiplying anything by $\alpha$ is to (a) expand/shrink by a factor of $r=|\alpha|$, and (b) rotate about the origin by an angle of $\theta$. Rotations and dilations are achieved by multiplication.

Now, suppose we want to rotate $T$ so it has an edge on the real axis, which will necessitate multiplying by something. We already know that $u\bar{u}=|u|^2\in\bf R$ for any $u\in\bf C$, so we can achieve our goal of rotating the triangle by multiplying the vertices by $\bar{w}_2$. Denote the rotated triangle $T'$, so that the vertices of $T'$ are $\bar{w}_2\cdot0=0$, $\bar{w}_2w_2=|w_2|^2$, and $\bar{w}_2w_1$. The base of this triangle has length $|w_2|^2$, and the height of this triangle (the vertical distance of the third vertex from the real axis) is ${\rm Im}(\bar{w}_2w_1)$ Hence ${\rm Area}(T')=\frac{1}{2}|w_2|^2{\rm Im}(\bar{w}_2w_1)$.

The issue at this point is that multiplying by $\bar{w}_2$ didn't *just* rotate the triangle $T$, it also *stretched* it. Luckily, multiplication stretches the complex plane uniformly - all lengths are scaled by a given factor. In particular, since lengths are stretched by a factor of $|w_2|$, areas must be stretched by a factor of $|w_2|^2$. Hence ${\rm Area}(T')=|w_2|^2{\rm Area}(T)$. Solving for ${\rm Area}(T)$ yields the conclusion.