I can’t believe I made this mistake 😂

Can you spot what this math professor did wrong?

[mudkip_btw]

Always be careful when deriving fast ;p

[JSSTyger]

Minus 30 points for that mistake. You cant have fractions of a pi m.

[ojasdeshpande]

I heard "can't" and "can" interchangeabled and was confused lol

[DoctrinaMathVideos]

I always tell my precalculus students that you can if you are working with complex numbers which we have already introduced towards the beginning of the course.

[physicsdemonstrations]

I did this same mistake in an interview when I was asked to find out if i^i is real or imaginary. Man I was embarrassed 🤦🏻, had a laugh. I did get selected though

[anshumanagrawal]

"Multiples of myself"
Lol that's smart

[diogoprudente]

Pi/2 Shouldnt it be just pi because of the arg.
I see e^i(pi/2)=i.
Pls correct me

[luizmiguel]

For whom didn't understand, the identity is actually "e^(iπ)=-1". "e^[i(π/2)]=i"...

[kayeassy]

It's exp(iπ) = -1
We all make mistakes.

[מיכאלקונטרוביץ]

Just one minute...it is not a course in complex analysis...in a regular calculus of real functions you can't do it...And if you decide to go into complex, then you have to explain the theory of multivalued functions...Dr Peyam, you right if you want to get the infinite set of values of ln of minus one. Also, for my knowledge, some complex analysis students, especially phisicists or engineers, do not really understand the difference between defining the function as a multivalued object, and defining its analytic branches...or defining some branch which value is defined, but the function is not analytic (actually not continuous). It is a tricky one

[danielrosado]

lmao 😂 that’s me every time I do complex analysis

[jimbyers]

I appreciate how you drive home that certain, innocent expressions, like "-1", have other representations (e.g., exp(iπ)) that are very helpful in certain contexts. I wonder if there is a name for such or, better, a list of some. Thank you!

[MathFromAlphaToOmega]

A nice application of taking logs of complex numbers is letting t=e^(ix) in
-log(1-t)=t+t^2/2+t^3/3+...
to get


Then the imaginary part of the left side is the negative of the argument of 1-e^(ix). That ends up being (pi-x)/2 as long as 0<x<2pi. So taking the imaginary part of both sides gives the Fourier series

[AarushLanjharia]

e^(i*pi)=-1; e^(i*pi/2)=i (For those of you unaware of euler's identity)

[marshallbruce]

That should be e to the i pi not by two.

[brettstafford]

Wait if ln(-1)=-ln(-1), then ln(-1)=0?

[gurkiratsinghtha]

It should be iπ not iπ/2.
Fun fact:- e^iπ/2 = i

[zerocks]

Isn‘t it e^ipi not pi/2 or am i being stupid rn

[dougr.]

Factors of too, two and to are important

[ganitaddhyayangrih]

Your answer is wrong because the value of ln(-1) is iπ not iπ/2.