For a couple days now, I have been thinking about the graph of the real and imaginary parts of complex functions, for example: $$f(z)=z^2=(x+iy)^2=x^2-y^2+i(2xy)$$
I found that if I set, for any constants c$$x^2-y^2=c_1, 2xy=c_2$$
Then if I zoom in more and more on the intersections of these two curves, the angle at which they intersect appears to always approach a right angle. I have tried this also for sines, cosines, exponentials, and other elementary functions and found the same result, but I am not sure what is actually going on here. Why is this happening? There must be something more fundamental about the nature of these functions that I am missing.
I have done some research into conformal maps that show this phenomenon for many, many values of c, but I don't believe I have quite developed the mathematical maturity for complex analysis quite yet. I am coming at this from the perspective of a motivated high school senior with knowledge of multivariable calc and some linear algebra/differential equations.
If anyone could provide me some insight or direction into, more generally, why this happening and what I am missing, I would greatly appreciate it!
Best Answer
This happens specifically with differentiable complex functions. I have a kind of intuitive explanation of this phenomenon.
First, as I hope you know, every complex number can be written in polar form, which looks like this $$r(\cos(\theta) + i\sin(\theta))$$ for some real $r \ge 0$ and $\theta \in \Bbb{R}$. Multiplying complex numbers in polar form is nice and intuitive. It's easy to verify that: \begin{align*} &r_1(\cos(\theta_1) + i\sin(\theta_1)) \times r_2(\cos(\theta_2) + i\sin(\theta_2)) \\ = \, &r_1 r_2(\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)). \end{align*} What this means is, if we are multiplying a complex number $z$ in general by a specific complex number $r(\cos(\theta) + i\sin(\theta))$, then all we have to do is:
Note that both of these operations preserve lines and don't change angles. If we start with two lines in the complex plane intersecting at an angle $\alpha$, then multiplying every complex number in these lines by $r(\cos(\theta) + i\sin(\theta))$ will result in two (rotated) lines, still intersecting at an angle of $\alpha$.
Now, let's say we have a complex function $f$ which is differentiable at some complex number $z_0$. Essentially what this means is that the function $f$ behaves like its linearisation: $$L(z) = f'(z_0)z + f(z_0)$$ for $z$ close to $z_0$. All that $L(z)$ does is multiply $z$ by a fixed complex number $f'(z_0)$ (which just scales and rotates), then adds a fixed complex number $f(z_0)$, which just shifts the picture without changing angles. So, $L$ will also not affect angles!
And since $f(z) \approx L(z)$ for $z$ around $z_0$, this means that, if two smooth curves meet at $z_0$ and their tangent lines meet at an angle $\alpha$, then whatever curves $f$ maps them to will have their tangents meet at angle $\alpha$.
So, what about your particular question? You have a complex function $f(a + ib) = u(a, b) + iv(a, b)$, which we will want to be differentiable. You are looking at the curves $u(a, b) = c_1$ and $v(a, b) = c_2$. The curve $u(a, b) = c_1$ is the set of points which will map to the vertical line $\operatorname{Re} z = c_1$. Similarly, $v(a, b) = c_2$ is the curve that will map to the horizontal line $\operatorname{Im} z = c_2$.
So, if the two curves intersect at $z_0$, and their tangents form an angle $\alpha$, then because $f$ is differentiable at $z_0$ (as it is everywhere else), the image of these curves must also form an angle $\alpha$. But, as we discussed, the image of these curves are horizontal and vertical lines, which are perpendicular! So, we must have $\alpha = \pi/2$. That is, if we look very closely at intersections between the curves, they must occur at right-angles.