Thanks for reading!
In this link...
https://betterexplained.com/articles/intuitive-understanding-of-eulers-formula/
...Euler's formula is explained pretty well.
I understand that multiplying a complex number by $e^{i} = (1+\frac{i}{n})^{n}$ as we let $n \rightarrow 0$ means that we're adding infinitely small copies of the complex number at $90^0$ from its tip $n$ times.
Each time we multiply the complex number by $(1+\frac{i}{n})$, the angle grows by $\textrm{arctan}(\frac{1}{n})$ radians, and the length of the complex number gets scaled by $\sqrt{1+\frac{1}{n^2}}$.
However, there's one part I don't really understand.
Due to each multiplication of some complex vector $z$ by $(1+\frac{i}{n})$, both the rotation and scaling are becoming insignificant as we let $(n \rightarrow \infty)$.
But, why is it that the rotation becomes "less insignificant" than the scaling does, so that as $n$ becomes really big, we can say that multiplying by this complex number is only a rotation, but not a scaling at all?
I'm looking for an intuitive answer, geometric and with pictures would be best.
I know that the limit of $\textrm{arctan}(\frac{1}{n})$ as $n \rightarrow \infty$ is $\frac{1}{n}$ radians, so that rotating the complex vector $n$ times corresponds to a total rotation of $1$ radian while the limit of $\sqrt{1+\frac{1}{n^2}}$ as $n \rightarrow \infty$ is $1$, which means we'd be scaling the vector by $1$, but $n$ times, which means its length doesn't get changed at all.
But, for some reason, that's not helping me.
Is there a more intuitive way to understand this? A more intuitive way to understand why although both the rotations and the scalings of our complex number are becoming infinitely small, in the limit the rotations still matter, while the scalings don't?
Thanks!