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A simple proof that e^(pi*i) = -1

I'll be honest, I only wrote this to test how well KaTex works

But this proof is quite neat, its something I learnt in high school and has stuck with me since.

I remember being curious as to how three numbers, Euler's number, pi and i, two which are transcendental and one being imaginary could give -1.

Proofofeπ∗i=−1z=cos⁡θ+i∗sin⁡θdzdθ=−sin⁡θ+i∗cos⁡θdzdθ=i(i∗sin⁡θ+cos⁡θ)dzdθ=i∗z∫1zdz=∫i∗θdθlog⁡ez=i∗θ+cei∗θ+c=zz=cos⁡θ+i∗sin⁡θWhenθ=0z=1sotheconstantis0..˙ei∗θ=z..˙cos⁡θ+i∗sin⁡θ=ei∗θWhenθ=π−1+i∗0=ei∗πei∗π=−1Proof\enspace of\enspace{e}^{\pi*i} = -1 \\ z=\cos{\theta} + i*\sin{\theta} \\ \frac{dz}{d\theta}=-\sin{\theta} + i*\cos{\theta} \\ \frac{dz}{d\theta}=i(i*\sin{\theta} + \cos{\theta}) \\ \frac{dz}{d\theta}=i*z \\ \int{\frac{1}{z}}{dz} = \int{i*\theta}{d\theta} \\ \log_e{z}=i*\theta + c \\ e^{i*\theta + c} = z \\ z=\cos{\theta} + i*\sin{\theta} \\ When\enspace \theta=0 \enspace z=1 \enspace so \enspace the \enspace constant \enspace is \enspace 0 \\ \dot{.\hspace{.075in}.}\hspace{.5in} e^{i*\theta} = z \\ \dot{.\hspace{.075in}.}\hspace{.5in} \cos{\theta} + i*\sin{\theta}=e^{i*\theta} \\ When \enspace \theta=\pi \\ -1 + i*0 = e^{i*\pi} \\ e^{i*\pi} = -1

Opinions on KaTex

It is pretty mediocre, I think perhaps compiling my LaTex into a png might have been a better idea. KaTex misses a lot of functions, and of course, you cannot import any existing LaTex packages.

Initially, I was thinking of doing a post on Lattices and ordering but I was limited by KaTex.

Perhaps this post will come eventually still. I am interested in writing about compiler optimisations (data flow analysis) and concensus algorithms, both of which rely on lattices. I largely want to do this to improve my own confidence on these two topics.