e to the pi i for dummies

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Recommends Mark S. · 2016-06 · Mathematics Stack Exchange · on “Simple Proof of the Euler Identity $\exp{i\theta}=\cos{\theta}+i\sin{\theta}$”

I have two favorite arguments that we should have $\exp (i\theta)=\cos \theta +i\sin \theta$ for real $\theta$ . The first is closely related to Mathologer's video e to the pi i for dummies [this video] , and the second is discussed in slightly more detail in II.2 “Moving Particle Argument” in Visual Complex Analysis [LINK] . Finally, I conclude with a summary of how Euler did it, from How Euler Did It [LINK] by Ed Sandifer for MAA Online. 1. Limit Argument Many high school students are aware that $e=\displaystyle{\lim_{n\to\infty}}\left(1+\dfrac{1}{n}\right)^n$ . For real $r$ , some may be acquainted with the fact that $e^r=\displaystyle{\lim_{n\to\infty}}\left(1+\dfrac{r}{n}\right)^n$ . We can

Recommends Mirek Olšák · 2017-02 · Mathematics Stack Exchange · on “Geometric interpretation of complex number.”

Complex number multiplication (and exponentiation) has a geometric interpretation. It is described for instance in this video [this video] . When you know that, the problem becomes just a problem of euclidean geometry and you should try to avoid algebra as much as possible. In general for euclidean geometry problems, I can recommend tools for playing with them, especially GeoGebra [LINK] . For this case choose point A and select "complex number". Create it by clicking to the plane. Activate "Input" on the left. Write "z_1"right arrow"^2"enter. Then similarly create z^3 and join them with the polygon tool. Then use the "mouse cursor tool" to move the original point $z_1$ and see what happens. Then

Recommends Ben Grossmann · 2017-02 · Mathematics Stack Exchange · on “Intuition for $e^{ix}$ lies on the unit circle.”

There are some videos that explain it nicely. See this one from 3Blue1Brown [LINK] or this one from Mathologer [this video] .

Recommends Ben Grossmann · 2017-06 · Mathematics Stack Exchange · on “False proof of $\pi=0$”

Here's a quick way to summarize the essence of your paradox: $$ e^{2 \pi i} = e^0 \overset{?}\implies 2 \pi i = 0 $$ The first equation is true, the second is not. The problem is that the log is no longer uniquely defined. If it were defined in that sense, then taking the "log of both sides" would take us from the first equation to the second. With real numbers, we can only have $e^x = e^y$ when $x = y$. This is not true if $x$ and $y$ are complex. Since you are learning about complex numbers for the first time, it is worth noting that for real numbers $x$, we have Eulers formula: $$ e^{xi} = \cos(x) + i \sin (x) $$ Notably, both cosine and sine are periodic, so this function will be periodi

Cites Skilly · 2018-03 · Mathematics Stack Exchange · on “Euler's Formula explained with exponential function”

Why is $$ e^x = \lim_{n \to \infty} \left( 1+ \frac{x}{n} \right)^n $$ also true for an imaginary $x$? Per definition, $$ e = \lim_{n \to \infty} \left( 1+ \frac{1}{n} \right)^n $$ so $$ e^x = \lim_{n \to \infty} \left( 1+ \frac{1}{n} \right)^{nx} $$ and $$ e^x = \lim_{n \to \infty} \left( 1+ \frac{x}{nx} \right)^{nx} $$ I understand now that we can substitute $n$ for $n*x$ as both approach $\infty$, but why does this also work for $x=i$, where $\lim_{n \to \infty} \left(n*i\right) $ would approach $\infty*i$, which is all the way to the top of the complex plane instead of to the right? $$ e^i = \lim_{n \to \infty} \left( 1+ i × \frac{1}{ni} \right)^{ni} $$ (This question came to me at 7:47

Cites Pineapple Fish · 2022-04 · Mathematics Stack Exchange · on “How to find $\arg(z)$ and $|z|$?”

This is just a starter to actually answer the question, I'd be willing to have a longer discussion in comments or chatroom; $\left(\cos(\frac{\pi}{5})+i\sin(\frac{\pi}{5})\right)^{15}(3-3i)^{20}=(e^{i\frac\pi5})^{15}\left(3\sqrt{2}e^{-i\frac\pi4}\right)^{20}=e^{3\pi i}(3\sqrt{2})^{20}\left(e^{-i\frac\pi4}\right)^{20}=e^{3\pi i}\cdot3^{20}\cdot\sqrt{2}^{20}\left(e^{20\cdot-i\frac\pi4}\right)=e^{3\pi i}\cdot3^{20}\cdot\sqrt{2}^{20}\left(e^{-5\pi i}\right)=3^{20}\cdot\sqrt{2}^{20}e^{-2\pi i}$ but note that $3^{20}\cdot\sqrt{2}^{20}$ , but again by laws of exponents you can simplify further because $\sqrt{2}=(2)^{\frac12}$ , so $3^{20}\cdot\sqrt{2}^{20}=3^{20}\cdot\left((2)^{\frac12}\right)^{20}