Inverse of a complex function In the real numbers, the inverse of a function could be visualized as a reflection across the line $y=x$. Is there an equivalent reflection in the complex plane?
 A: The same reflection works, but in $\mathbb{C}^2$, rather than $\mathbb{R}^2$. The map that sends a graph of a function to the graph of its inverse is
$$(x, y) \mapsto (y, x),$$
regardless of which sets $x$ and $y$ belong to ($\mathbb{R}$, $\mathbb{C}$, or any other set). In $\mathbb{R}$, we can visualise this as a reflection. In $\mathbb{C}$, it's a little trickier to actually visualise, due to the four real dimensions, but we can still define a reflection in the line $\lbrace (x, x) : x \in \mathbb{C} \rbrace$, using inner products.
First, define a projection map. Recall the complex inner product
$$\langle (a, b), (c, d) \rangle = a \overline{b} + c \overline{d}.$$
Let $L = \lbrace (x, x) : x \in \mathbb{C} \rbrace$. We can define the orthogonal projection of a point $(x, y)$ onto $L$ by
$$P_L(x, y) = \frac{\langle (x, y), (1, 1) \rangle}{\langle (1, 1), (1, 1) \rangle} (1, 1) = \frac{x + y}{2}(1, 1).$$
Then, the reflection in $L$, $R_L$, is defined by adding the vector from $(x, y)$ to $P_L(x, y)$ onto $P_L(x, y)$ (this makes some sense if you draw a quick picture on the real plane). That is,
$$R_L(x, y) = 2P_L(x, y) - (x, y) = (x + y, x + y) - (x, y) = (y, x).$$
So, $R_L$ is the map we're looking for.
Note that this is basically the same way you'd use to show the real case!
