Thats awesome! Although I'm not a english speaker, I got everything! Please, make the next part! Sigma

hypotenoose

Thanks for this clear video lesson on the application of integration to area. How about one on volume please?

Can you please make one more video on this topic and bring some easy questions like this for more understanding.

I thinks that squaring both sides gives extra solution

Wait, his may be a dumb question but how do you get one here, 6:45

Let’s not confuse a misleading problem with a “clever” problem. You suffer a loss of credibility if not integrity. Do better next time.

xis 13or9

sin(x)=2 x=i ln(i-3+2i)+360n For those who dont get it,a horizontall bar is used for square root.

log(2+sqrt(3)) = -log(2-sqrt(3)), so the plus minus could have been before the log instead, or the minus could be flipped to plus.

3:40 i dont like the way those rectangles stick out Is there a way to Destroy them Hmmmmm Its forming a triangle thingy but The hypotenuse goin curvy curvy Wierd

Can you do something like this for trigonometry too plsssss

Easy to understand Very good

OH MY GOD FINALLY SOMETHING THAT ISN'T "Random cool stuff" I dont dislike random cool stuff but THIS IS GREAT!!! i sub

It's very useful Thank you.....

Awesome video... Easy to understand ❤❤

The problem is the fact that you toy with imaginary so effortlessly as if it were nothing more than a secondary school problem. Using powers and logs of imaginary and negative real numbers is dangerous as many paradoxes arise. The same laws you write for real numbers such as (a^b)^n = a^bn does not work for imaginary numbers. Also raising by powers of non integer numbers isn't nice since there can be multiple answers. Many of the integer solutions you found were incorrect as they were complex solutions. Also as a final thing, you can't just "take the natural log" of a complex number, because the answer is not really like the single line graphs you get on the real plane. (I just started Calculus please correct me if I'm wrong, I'm still learning😅)

This is one of your best video till date! It was so easy to follow and I understood everything! 🔥🙏

when π=0 e^iπ +1=0

m^3+/-m^2+11m-30=0 , (m-2)(m^2+2m+15)=0 , 1 -2 2 -4 15 -30

Sir what is angle?? Exact definition

Maybe the AI narrating this figured it out by itself!

This is not better than the usual formula !

Bro squared the equation and literally introduced ANOTHER SOLUTION to the given equation LMAO.

2x2-20x +50 please solve it by this method

Skibidi😮😮 This method is cool..Thanks sir

Sir what if it would be downward graph

LYING about the dimensions made me angry. Claiming to have a "final answer" was also a lie, as it was known the problem had no solution. (Confessing the second lie does not constitute a "final answer.") I won't waste my time on any other of your posts.

Calling natural logarithm log...okay.

ig the comments are saying this isn't true, thank god

(e^a)^b does not equal e^(ab) unless a is real or b is an integer thus (e^(2 pi I + 1))^(2 pi I + 1) e e^[(2 pi I + 1)(2 pi I + 1)] ( e means not equal)

It's good for quick answer। (-2.75,16.01i), (-2.75,-16.01i). Thank you for summarising।😊😊

I use a method I learnt in school if it will factorise. Doesn’t work so well if there are irrational roots! A x^2 + B x + C Multiply A by C to get AC. If AC is +ve you want factors of AC that add up to B. If AC is -ve then factors of AC that differ by B. So take x^2 - 22 + 117. AC = 117 so we want factors of 117 that add up to 22. 117 = 3 x 39 = 9 x 13. 3+39 = 42, too big; but 9 + 13 =22 So we have (x-9)(x-13). Takes 2 or 3 seconds once you are familiar with the process!

That last step has (2pi)^2=0, hence 2pi is 0 just as we'd expect. We want 2pi to be 0, not pi itself.

Solutions are 9 and 13 because 117 = 9x13 and 22 = 9 + 13 (Written before watching the video) 😅

No formula. He proceeds to use Po-Shen Loh formula.

I'd use factorization or quadratic formula instead. Just click formula 01 on the calculator and the results appear

as some people said, the only mistake on this "proof" (which is enough to get you to the contradiction) is that (a^m)^n = a^{mn}. This identity only holds for non negative real values of m,n. Also, raising complex numbers to the power of non integer values is also quite dangerous (e is real but e^{2pi+1} is not). By dangerous i mean its not well defined, anyway, moral of the story is dont walk on graveyards. dont mess with what you dont know how to mess with

This is just the _Clausen paradox_ published by Thomas Clausen in 1827. See the English Wikipedia article on _Exponentiation_ for an explanation. Some of the identities for powers and logarithms used in this video, while valid for positive real numbers, do not apply to complex numbers, no matter how complex powers and complex logarithms are defined as single-valued functions.

“We don’t have to rely on memorization.” Um… that’s not true, you’re just memorizing two formulas for S and M instead of one quadratic formula. It’s actually faster to either use the quadratic formula straight away. (Or if you’re specifically looking for rational roots use the Rational Root Theorem to quickly get the answer.)

Just split the middle term as the sum of 'b' and the product of the two split terms must be equal to 'a*c'.

This can all be avoided by using the complete form e^(i * pi * (2n + 1)) = -1

Because of how the "presenter" was talking, I didn't absorb any of it. I'll to watch it a few more times to get it.

As an Indian who can do quick calculations in mind only, I approve of this method for larger numbers. Helpful!

The power rule you have stated (which is (a^b)^c = a^(bc)) is only true for non negative real numbers, it does not hold for negative or complex numbers. its like using the fact that (a+b)^2 = a^2 + b^2 on any number despite the fact that it is only true when a or b = 0 also you overcomplicated this to make it sound smarter, if you wanted to extract a contradiction by using rules that dont hold up when using complex numbers you couldve just done e^itheta = cos(theta) + isin(theta) e^i2pi = cos(2pi) + isin(2pi) = 1 e^2ipi = 1 ln(e^2ipi) = ln(1) (The mistake in this case is here as you cannot take the natural logarithm of both sides with complex numbers since unlike in the real numbers where the natural logarithm is a one to one function, in the complex numbers the natural logarithm is not a one to one function as for example ln(1) can equal n2ipi where n is an integer, so it can be 0 or 2ipi or 4ipi etc...) 2ipi = 0 pi = 0

Mast hai

Euler's Not *oilers*

Well, if it breaks all the rules, then it can't possibly be correct 😂.

This method is nothing but using the quadratic formula in pieces.

Taking the natural log of (-1)^2 equals 2ln(-1) which isn’t a real solution, it does not equal 0. When you squared both sides, you made a new solution that doesn’t exist