TITLE: Help understanding $e^{it}=\cos t+i\sin t$ by way of matrices and vector fields
QUESTION [6 upvotes]: I was brushing up on my complex arithmetic in preparation for a class in ODE's this semester and I found myself looking at Exercise 2.7.5 in Introduction to Complex Analysis for Engineers by Michael Alder, which reads

The exponential function is a procedure for turning vector fields into flows; if you take the vector field which is given by$$V\begin{bmatrix}
x \\
y
\end{bmatrix}=\begin{bmatrix}
0 & -1\\
1 & 0
\end{bmatrix}\begin{bmatrix}
x \\
y
\end{bmatrix}$$
  you call the matrix $A$ and then the flow is given as $e^{tA}$.
  [...]
  Draw a picture of the vector field. Identify the matrix as a complex number. Deduce that $e^{it}=\cos t+i\sin t$ is little more than the observation that exponentiation is about solving ODE's by Euler's method taken to the limit.

I would like very much to understand this very well. I've actually done most of it and perhaps the problem is that I haven't actually taken the ODE's class yet, but I've read ahead enough to understand most of what's being said.
I drew the vector field (by hand) and got some lovely circley looking things. When he says, "identify the matrix as a complex number," I understand that he is referring to the fact that in the book he defines a complex number $a+bi$ to be the matrix $$\begin{bmatrix}
a & -b\\
b & a
\end{bmatrix}$$and so $A$ is $i$. I also managed to do the exponentiation $e^{tA}$ and got
$$\begin{bmatrix}
\cos t & -\sin t\\
\sin t & \cos t
\end{bmatrix}$$
Which is, of course, the complex number $\cos t + i\sin t$.
So so far so good, I've shown that $e^{it}=\cos t + i\sin t$.
I'm just having problems understanding the last little bit, and maybe that's cause I haven't taken the ODE's class yet, but I see that there are tangent lines to radiuses of circles somewhere in there since the vector field makes tangent lines to circles around the origin and $e^{At}$ ends up being the rotation matrix with angle $t$, so we have some notion of a radius rotating around the origin somewhere? How does this relate to Euler's method for solving ODE's? Is the idea that there is an ODE which produces that vector field as a direction field, and $e^{it}$ gives solutions? I get a little bit lost at this point, maybe someone can help me finish putting the pieces together.

REPLY [3 votes]: To be explicit, we have the ODE
$$\frac{d}{d\tau}z(\tau) = iz(\tau)$$
which we want to integrate over $\tau \in [0,t]$. Divide the interval into $n$ steps of length $t/n$ each. Euler's method gives
$$z_{k+1} = z_k + \frac tn\cdot iz_k = \left(1 + \frac{it}n\right) z_k$$
and so
$$\quad z_n = \left(1+\frac{it}n\right)z_{n-1} = \left(1+\frac{it}n\right)^2z_{n-2} = \cdots = \left(1+\frac{it}n\right)^nz_0.$$
Does that look familiar?