This showed up as part of an integration problem I was helping a calculus student with. They asked whether or not it was was possible to get exact solutions to which I responded "not in any way that will help."
I then attempted to determine the solution set.
The first thing to note is that $0$ is a solution. With that out of the way, we can look for the nontrivial solutions.
$$\begin{align} \frac{e^{iz^2}-e^{-iz^2}}{2i}=rz^2&\implies e^{iz^2}-e^{-iz^2}=-2irz^2\\ &\implies\sum_{n\in\Bbb N}\frac{(iz^2)^n}{n!}-\sum_{n\in\Bbb N}\frac{(-iz^2)^n}{n!}=-2irz^2\\ &\implies\sum_{n\in\Bbb N}\frac{(iz^2)^n-(-iz^2)^n}{n!}=-2irz^2 \end{align}$$
Noting that $(iz^2)^n=(-iz^2)^n$ when $n$ is even, so suppose that $n=2k+1$ for some $k\in\Bbb N$. We then have
$$\begin{align} (iz^2)^n-(-iz^2)^n&=(iz^2)^{2k+1}-(-iz^2)^{2k+1}\\ &=ii^{2k}z^{4k+2}-(-i)(-i)^{2k}z^{4k+2}\\ &=i(-1)^kz^{4k+2}+i(-1)^kz^{4k+2}\\ &=2i(-1)^kz^{4k+2} \end{align}$$
substituting this into the previous formula and assuming $z\ne0$, we get
$$\begin{align} 2i\sum_{k\in\Bbb N}\frac{(-1)^kz^{4k+2}}{(2k+1)!}=-2irz^2&\implies\sum_{k\in\Bbb N}\frac{(-1)^kz^{4k+2}}{(2k+1)!}=-rz^2\\ &\implies-\frac1r\sum_{k\in\Bbb N}\frac{(-1)^kz^{4k}}{(2k+1)!}=1\\ &\implies 1+\frac1r\sum_{k\in\Bbb N}\frac{(-1)^kz^{4k}}{(2k+1)!}=0 \end{align}$$
Could I have accomplished the same thing by looking in literally any calculus textbook and finding the Taylor series for the sine function? Yes. Yes, I could have.
Now, since $z\ne0$ by assumption, we know that every term of the summation is nonzero. From here it is easy to check that the series does not reduce to a polynomial. Hence, every nonzero solution is transcendental. And this is pretty much the only reason for involving the Taylor series.
Noting that $f,g:(0,\sqrt\pi)\to\Bbb R;f(x)=rx^2,g(x)=\sin(x^2)$ are analytic, bounded, increasing/decreasing, etc. we can prove that there is at least one real solution in the interval $(0,\sqrt\pi)$. Hence, there is at least one positive real solution. Call the least such solution $z_0$.
Next, suppose that $z$ is any solution to our equation. Obviously, $(iz)^{4k}=(i^4)^kz^{4k}=z^{4k}$. Hence,
$$1+\frac1r\sum_{k\in\Bbb N}\frac{(-1)^k(iz)^{4k}}{(2k+1)!}=1+\frac1r\sum_{k\in\Bbb N}\frac{(-1)^kz^{4k}}{(2k+1)!}=0$$
In other words, $iz$ is a solution wherever $z$ is.
We now have five solutions in total: $0$, $z_0$, $iz_0$, $-z_0$, and $-iz_0$.
This is where I'm stuck. It's obvious that there are at most finitely many positive real solutions (it might be possible to give the exact number in terms of $r$, but I'm not doing that without a reason), and subsequently finitely many real and pure imaginary solutions (these being the product of real solutions with powers of $i$). But what about complex numbers off of the real and imaginary lines?
Suppose $z=x+iy$, $x,y\ne0$, then
$$\begin{align} \frac{e^{iz^2}-e^{-iz^2}}{2i}&=\frac1{2i}\left(e^{i(x+iy)^2}-e^{-i(x+iy)^2}\right)\\ &=\frac1{2i}\left(e^{i(x^2-y^2)-2xy}-e^{-i(x^2-y^2)+2xy}\right)\\ &=\frac1{2i}\left((\cos(x^2-y^2)+i\sin(x^2-y^2))e^{-2xy}-(\cos(x^2-y^2)-i\sin(x^2-y^2))e^{2xy}\right)\\ &=\frac1{2i}\left(\cos(x^2-y^2)e^{-2xy}+i\sin(x^2-y^2)e^{-2xy}-\cos(x^2-y^2)e^{2xy}+i\sin(x^2-y^2)e^{2xy}\right)\\ &=\frac1{2i}\left(\cos(x^2-y^2)(e^{-2xy}-e^{2xy})+i\sin(x^2-y^2)(e^{-2xy}+e^{2xy})\right)\\ &=\sin(x^2-y^2)\cosh(2xy)+i\cos(x^2-y^2)\sinh(2xy) \end{align}$$
From here,
$$\sin(x^2-y^2)\cosh(2xy)+i\cos(x^2-y^2)\sinh(2xy)=r(x+iy)^2=r(x^2-y^2)+i2rxy$$
This gets us the system
$$\begin{cases}\sin(x^2-y^2)\cosh(2xy)-r(x^2-y^2)=0\\\cos(x^2-y^2)\sinh(2xy)-r2xy=0\end{cases}$$
I have no idea how to solve this system, but I can see that the level sets given by the two equations intersect at several points. In fact, the points of intersection appear to occur in pairs to either side of the real or imaginary lines. From here I would guess that the set of solutions is countably infinite, with the frequency of solutions increasing with distance from the origin, but beyond that, I'm totally lost.
How do I determine the number of solutions, and can I characterize the set of solutions in any more detail?
How do I tag this? I don't think this counts as a calculus question, but there isn't a whole lot actual analysis going on here. I'm really just trying to solve the problem and not thinking too hard about what subject it falls under.