Recursive i

sorry, i assumed we were on the principle branch, and not "nitpicking for assholes"

that's what it says in my math book
you're not supposed to be ambiguous

insults aside, i will be serious with you.

ambiguity certainly is undesirable in math. however, relaxing your rigor to understand a concept is great first step. afterward, you can add back in the rigor and make your work crystal clear. the sad truth is that thinking purely rigorously gets extremely difficult once you are past the introductory (aka undergrad) level. you can expand you knowledge faster with the technique mentioned above.

Lmao it's fucking hilarious how these undergrad freshmen know-nothings here try to bring you down nitpicking every detail trying to appear smart. Great job OP on the discovery, keep it up! Math is beautiful. These are the type of things that convinced me to study math further and I hope you will find more of these nice discoveries.

[math] \displaystyle
f(x) = e^{-ix}(\cos x + i \sin x)
\\
f^{\prime}(x) = e^{-i x}(i \cos x - \sin x) - i e^{-i x}(\cos x + i \sin x)
\\
f^{\prime}(x) = e^{-i x}(i \cos x - \sin x) - e^{-i x}(i \cos x + i^2 \sin x) \equiv 0
\\
f^{\prime}(x) = 0 \;\;\; \forall \; x \in \mathbb{R}\Rightarrow f(x) \text{ is a constant}
\\
f(0) = e^{0}(\cos 0 + i \sin 0) = 1 \cdot(1+0) = 1 \Rightarrow f(x) = 1 \;\;\; \forall \; x \in \mathbb{R}
\\ \\
1 = e^{-ix}(\cos x + i \sin x) \Rightarrow e^{ix}=\cos x + i \sin x \;\;\; \forall \; x \in \mathbb{R}
[/math]

this is a fun one. i like the proof using the matrix exponential too.

yeeeeeeeeeeeeppppppppppp

>sqrt(4)=2

It's rare getting a compliment here.
Thanks

[math]\sqrt{4} = 2[/math]
[math]\sqrt{-1} = i[/math]
[math]\sqrt{1} = \iota[/math]

[math]\sqrt{\iota} \times \sqrt{-1} = i \sqrt{\iota }[/math]