i^i=x Taking logarithms i lni = lnx lni = -i lnx (1) Taking Euler's famous formula e^(i pi)= -1 And taking the sqrt. e^(i pi/2) = i Substituting in eq 1 (i pi/2) = -i lnx You get rid of the i. And obtain: x=e^(-pi/2)
One of the most difficult challenges for any teacher is to make the video worthwhile for a broad range of student backgrounds. You've got an uncanny knack for doing this well.
Very simple Question : i^i=e^(ln(i)*i)=e^(ln(sqrt(-1)*i) e^(1/2*ln(-1)*i) We know : e^(pi*i)=-1 Placement; e^(1/2*ln(e^(pi*i))*i)=e^(1/2*i²*pi)= 1/e^(pi/2) that is a real number 😁
it is true from the perspektive of theresult👍. still, I wonder if we can breake the none negative law inside ln(x) becasue you mentioned ln(-1)in your process.
I think a good Step One would be to define f (w) = z^w in the first place, for a given z. [That would involve the logarithm, which in passing helps to grasp the multivalued nature of this function.]
Related note, the values e^(-(pi/2) + 2pii) aren't 'solutions,' they are just values of a function, for that input. Just like '4' is the value of the function x^2, at x = -2. It's not the 'solution' of something.
@@MathCuriousityhey, good question .. so "multivalued" is a thing where you put in a single value as an input, and you get two or more values out. "Square root" is an example, like how 25 -> 5 and -5 .. inverse trig functions are another example. And in a way, same idea : "I have a value N, and I need a value X with X^2 = N .. Or sine (X) = N ..
@@MathCuriousity oh, and for sure, also this : "A complex logarithm of a nonzero complex number Z, defined to be any complex number W for which e^W = Z .." en.wikipedia.org/wiki/Complex_logarithm
Today's question was pretty simple. Suggestion of a problem to be solved: sin(x) = 2. The solution exists (obviously, it is complex) and can be obtained analytically.
He pretty much sticks to simplifying expressions that could be used during a freshman calculus course. Especially the gotcha test questions that cull the herds. (:
If we consider Euler's identity for both e^ix and e^(-ix), and subtract the latter from the former, we get sin(x) = (e^(ix) - e^(-ix))/2. So 2 = (e^(ix) + e^(-ix))/2. Set u = e^ix and we get u + 1/u = 4, which gives u^2 - 4u + 1 = 0, whose solutions are 2 ± √3. Therefore x = -i.ln(2 ± √3).
@@RexxSchneider In fact, sin(x) = [e^(ix) - e^(-ix)]/2i . That is, you forgot an "i" in the denominator. For that reason, the final result is not correct.
@@walterufsc Yes, you're quite right. Thanks for the correction. The method remains unchanged so should read: u^2 - 4ui + 1 = 0, whose solutions are 2i ± √(-5) = (2±√5)i. Therefore x = -i.ln((2±√5)i). That can be further simplified to -i.ln(2±√5) -i.ln(i) which equals -i.ln(2±√5) - i(πi/2 + 2nπi) = -i.ln(2±√5) + π/2 + 2nπ where n ∈ ℤ.
Privet Nonna. That means minimum real x is 3.64417. Or i^i = 3.64417^3.64417. For k=2, x= 6.08754. For k=6, 13.7746. Whatever they all mean, is beyond me.
Another method may be starting from e^iπ = -1 so that iπ = ln(-1) = ln(i^2) = 2ln(i) -> π/2 = ln(i)/I -> ln(i) = iπ/2. Now we can write x = i^I -> ln(x) = iln(i) = i^2 * π/2= -π/2 -> e^ln(x) = x = i^I = e^(-π/2)
That occurred to me too, e^ipi = -1 but so is e^i*3*pi, e^i*5*pi, etc...leading to multiple answers for ln(-1). Does this mean i^i has multiple values?
Neatness is a key element to communicate ideas. As a teacher, explaining such an advanced topic, it would help if you had better penmanship skills. Learn to make an algebraic symbol x (two letter 'c's back to back), to distinguish it from the common multiplication sign. If my brain has to spend less time deciphering your untidy presentation, then I'll be better able to focus on your teaching.
Exponents in the complex world are multivalued. So 1^i has many solutions, 1 being just one of them. 1= e^(i2npi), n€Z 1^i = (e^(i2npi))^i = e^(i2npi*i) = e^(-2npi) For n=0 you get e^0 = 1. For n=1 you get e^(-2pi). For other n you get different solutions.
I've seen this answered as i^i = (e^i*pi/2)^i = e^-pi/2 ~= 0.2. What I don't understand is that following this reasoning, it is also (e^5*i*pi/2)^i = e^-5*pi/2.
If you take a real number and raise it to i, you get e to the quantity of i times the natural log of the real number, which is equal to the cosine of the natural log of the real number plus i times the sine of the natural log of the real number. This is how Euler’s formula works. It works for every complex number. Example four to the quantity of five plus three times i is equal to four to the five times e to the quantity of i times the natural log of four to the three. But here’s the question. If you take a complex number to a complex number, will the formula work? If it does work, here’s how you explain it with only math symbols: i^i = e^i ln i = cos(ln i) + i sin(ln i). But we can’t take the natural log of i. But oenr’s comment that is liked by the creator of this video says it’s real. And Josep Martí Carreras’s comment and SyberMath’s answer says it’s e^-pi/2. So the answer is absolutely e^pi/2.
please explain your point more clearly for a newb like me! I am a bit confused. Are you saying the author is wrong in this video!? Also How did he even get that initial equation in the beginning of the video. Where the heck does it come from?
great video. Can you explain Cauchy's theorems? I have never understood Complex analysis because it was never explained to me in College. We just memorized the results and used them in examples. I would love someone to explain it because I have no idea how Cauchy came up with any of it. The man must have been an abstract genius.
so is this solution a result of bad math or this solution is ACTUALLY the solution and the graph of x^x would intersect the solution if that graph was to be drawn in complex plane too.? someone elaborate please
Euler's identity: e^(ix) = cos(x) + i.sin(x) So: e^(-ix) = cos(-x) + i.sin(-x) = cos(x) - i.sin(x) Adding them, we get: e^(ix) + e^(-ix) = 2.cos(x). So cos(x) = (e^(ix) + e^(-ix))/2. When x = i, we have cos(i) = (e^(i.i) + e^(-i.i))/2 = (e^(-1) + e^(1))/2 = (e + 1/e)/2, which is a real number. Also equal to cosh(1) by definition.
You can't just multiple powers of i time i since they are complex numbers. That rule doesn't apply to complex numbers, even though you got lucky the answer came out correct. That rule only applies to real numbers. Please re-think is over and make the necessary correction to the steps. You are on the right track.
This could have been solved in a few seconds! Simply replace base i by e^i*pi/2 , raise it to power i then multiply the exponents to get -pi/2 Why such an elaborate video?
@@SyberMath no it is a complex number. Any number of the form a+ib is a complex number in which a is the real part and b is the imaginary part. An imaginary number is something that cannot be represented on a number line.
not real is not the same as imaginairy. as classification. for example. the number 10 is a part of the number 100 axioma's 1.real is not imaginairy 2.reality alowes imagination. therefor. hypothesis: 1.imagination is a part of reality bit not the whole reality, true or false ? so of 10 is a part of a bigger number. 100 things you dont imagine and are not part of reality can also excist. then negative 10 is not a 100 but 10 too the power of 10 is. so the question you ask is. does your imagination covers the whole reality ? or. do you have an accurate view on reality or do you intrepretate reality well, without observing it ? is that the line of your real question ? or are you missing something. like is the language too abstract too describe reality. (the answer on the last one is, yes, the language is merely a symbolicle representation of the real thing but sufficient enought to work with. ) or are you just saying hi...
your understanding of math should be measured by how far you can make it through this video before you give up trying to understand it and just watch to see what the answer is
Why in the world did you bring Euler's Identity into it? Completely unnecessary. Why didn't you write i = exp((i*Pi)/2) and then use your rules of exponents. Would have yielded the same result in less than 1 line.
Be careful when presenting graphs of solutions without proper identification of your graphs. You went from presenting the solution of i=cosx+isinx on the complex plane to presenting a solution of i^i=e^(-pi/2) on the real axis. Teaching complex numbers can be a bear ...
Let i^i = r[cos(a)+i*sin(a)] ---- (1) (i^i)^2=r^2[cos(2a)+i*sina(2a)] i^(i^2)= i^-1= 1/i= i/(i*i)= i/(-1)= -i= Then by comparing the real part and imaginary part, we can have the r and a. Sub r and a into (1), then problem solved. answer. 😊😊😊😊😊😊
пишут так что бы сбить с толку! 2:20 экспоненциальная запись это е в степени i * (фи), а если вместо фи используется х то и на рисунке должно быть х, но на рисунке пишут не х, не фи... а большую Тета! Вот зачем такие запутки? Написал х и на рисунке пиши х! А если не х то изначально пиши Тета и не морочь голову! 2:50 А объяснение синуса и косинуса через r, а потом в итоге выписано классическое определение синуса и косинуса (отношение определенного катета к гипотенузе)..... Жесть! Все сделано через левое плечо и правое колено! Жесть! синус есть синус, а косинус есть косинус! На кой тут r применили.... (я знаю что такое r но тут оно не к месту!) 3.00 а зачем писать нормальную форму комплексного числа для i и еще уточнять что мы имеем r=1 ? Вот на кой это? 3:30 ну да-да-да, 30 сек рассуждаем как комплексное число i будет отображаться на комплексной плоскости. Простите но это очевидно! (исходя из нормальной формы записи числа i) 3:50 записали i в тригоном.форме... а на 2:20 почему это нельзя было сделать? короче... до сих пор это все теория.... ладно... пусть так! 4:20 это можно было написать тогда когда экспоненциальную форму описывали т.е. 2 минуты назад.... (ок не ворчу......) 4:45 ну да.... только это одно действие и есть решение, а все остальное это теория..... которую должны знать те кто решают задачи с комплексными числами! (пол ролика теории.... нуладно! если автор так видит...) 5:20 вспоминает про период в два Пи.... мда.... печально! Вообще то это изначальное определение! И опускание этого факта в самом начале - это ошибка! короче! Баба Яга против! (ну не кошерно это!)
if i = -1 × real. if imaginairy is a negative real. there are a lot of "notreal" but they are not all the opposite of real. yhe words come from latin or greek. 1real 2.anti real 3.contra real 4.pseudo real 5.quasi real 6.beyond real 7.before real 8. areal 9.non real 10.not real 11.advanced real 12.not real yet 13.was real once but not know 14.is real for you but not for me 15.phantasy 16.imaginairy 17.like a dream 18.surreal (more then real) something like that. imaginairy is not that black and white or diabolic as presumpted. like if real state too imaginairy then if turning left state too going up... as ? instead of simple turning right. it is possible that imaginairy is more .. then reality.. although it is reality that makes imagination possible. if you ask me... everything is real. found out have a very efficient minimal perceptual view on the matter. senses are body specific. so pretty sure reality is more then can be perceived.
If i=e^i.pi/2 then (i)^i= (e^i.pi/2)^i Here folowing you will making a logic syntax error: You get power i in processin into e^i.pi/2 and resulting i^i= e^[(i^2).pi/2] You can not go like that!
Conclusion: if you take something imaginary and give it imaginary power, it becomes real.
Trascendental I might add
@@assassinosoldato92 Still real tho
And not only that, but anything that is real can still be complex.
Thank you for the feedback! This is incredibly unimaginary 😜
❤
i^i=x
Taking logarithms
i lni = lnx
lni = -i lnx (1)
Taking Euler's famous formula
e^(i pi)= -1
And taking the sqrt.
e^(i pi/2) = i
Substituting in eq 1
(i pi/2) = -i lnx
You get rid of the i. And obtain:
x=e^(-pi/2)
Nice.
but its not the only answer, there are infinity of them: x=e^(-p/2+2p*n) where n is integer
@@jemon2423 Of course. I always forget the 2pin term. 👍
@@jemon2423 why are there infinity of them! Where does 2pn come from?
One of the most difficult challenges for any teacher is to make the video worthwhile for a broad range of student backgrounds. You've got an uncanny knack for doing this well.
Thank you for the kind words! 💖🥰
So nice! You, sir, got me hooked on math again, almost 30 years after one lousy teacher ruined ir for me. Cannot thank you enough.
Glad I could help!
It wasn't your teacher's fault.
@@MrGrumbleguts wow, and you managed to write that, even without opposable thumbs!
@@MrGrumbleguts "YOU WERE THERE," a theme park or something
Very simple
Question : i^i=e^(ln(i)*i)=e^(ln(sqrt(-1)*i)
e^(1/2*ln(-1)*i)
We know : e^(pi*i)=-1
Placement;
e^(1/2*ln(e^(pi*i))*i)=e^(1/2*i²*pi)=
1/e^(pi/2) that is a real number 😁
it is true from the perspektive of theresult👍. still, I wonder if we can breake the none negative law inside ln(x) becasue you mentioned ln(-1)in your process.
Equivalent reasoning:
x = i^i
=> x^i = i^(-1) = 1/i = i/i² = -i
If x = e^θ:
x^i = e^(iθ) = cosθ+isinθ = -i
=> cosθ = 0 and sinθ = -1
=> θ = 3π/2 + 2kπ
=> x = e^(3π/2 + 2kπ)
I think a good Step One would be to define f (w) = z^w in the first place, for a given z. [That would involve the logarithm, which in passing helps to grasp the multivalued nature of this function.]
Related note, the values e^(-(pi/2) + 2pii) aren't 'solutions,' they are just values of a function, for that input. Just like '4' is the value of the function x^2, at x = -2. It's not the 'solution' of something.
are you referring to what Jmart did above in his comment? What does multivalued nature mean? Can you explain further? Thanks so much!
@@MathCuriousityhey, good question .. so "multivalued" is a thing where you put in a single value as an input, and you get two or more values out. "Square root" is an example, like how 25 -> 5 and -5 .. inverse trig functions are another example. And in a way, same idea :
"I have a value N, and I need a value X with
X^2 = N ..
Or
sine (X) = N ..
@@MathCuriousity more info :
en.wikipedia.org/wiki/Multivalued_function
@@MathCuriousity oh, and for sure, also this : "A complex logarithm of a nonzero complex number Z, defined to be any complex number W for which e^W = Z .."
en.wikipedia.org/wiki/Complex_logarithm
Today's question was pretty simple. Suggestion of a problem to be solved: sin(x) = 2.
The solution exists (obviously, it is complex) and can be obtained analytically.
He pretty much sticks to simplifying expressions that could be used during a freshman calculus course. Especially the gotcha test questions that cull the herds.
(:
If we consider Euler's identity for both e^ix and e^(-ix), and subtract the latter from the former, we get sin(x) = (e^(ix) - e^(-ix))/2.
So 2 = (e^(ix) + e^(-ix))/2. Set u = e^ix and we get u + 1/u = 4, which gives u^2 - 4u + 1 = 0, whose solutions are 2 ± √3. Therefore x = -i.ln(2 ± √3).
@@RexxSchneider In fact, sin(x) = [e^(ix) - e^(-ix)]/2i . That is, you forgot an "i" in the denominator. For that reason, the final result is not correct.
@@walterufsc Yes, you're quite right. Thanks for the correction. The method remains unchanged so should read:
u^2 - 4ui + 1 = 0, whose solutions are 2i ± √(-5) = (2±√5)i. Therefore x = -i.ln((2±√5)i).
That can be further simplified to -i.ln(2±√5) -i.ln(i) which equals -i.ln(2±√5) - i(πi/2 + 2nπi) = -i.ln(2±√5) + π/2 + 2nπ where n ∈ ℤ.
@@RexxSchneider That was very nice thank you
GOOD EVENING SIR
The expression is > 1÷5 when k=0 and < 1÷5 when k > or = 1 and ＞1÷5 as k is less than or equal to -1
Thank you
For k=1 i^i = e^(1.5*π) = 111.318
When dividing the length of the equator by 360 degrees, we get 111.319 km
Whatever that means? 😄
Privet Nonna. That means minimum real x is 3.64417. Or i^i = 3.64417^3.64417. For k=2, x= 6.08754. For k=6, 13.7746. Whatever they all mean, is beyond me.
Privet. Now you have to find answer...
Another method may be starting from e^iπ = -1 so that iπ = ln(-1) = ln(i^2) = 2ln(i) -> π/2 = ln(i)/I -> ln(i) = iπ/2. Now we can write x = i^I -> ln(x) = iln(i) = i^2 * π/2= -π/2 -> e^ln(x) = x = i^I = e^(-π/2)
That occurred to me too, e^ipi = -1 but so is e^i*3*pi, e^i*5*pi, etc...leading to multiple answers for ln(-1). Does this mean i^i has multiple values?
@@ianboard544 yes
@@jemon2423 This is fascinating. Going to have to chew on this for a while.
Thank you so much. Really happy to learn something new today. 😄😃
Neatness is a key element to communicate ideas. As a teacher, explaining such an advanced topic, it would help if you had better penmanship skills. Learn to make an algebraic symbol x (two letter 'c's back to back), to distinguish it from the common multiplication sign.
If my brain has to spend less time deciphering your untidy presentation, then I'll be better able to focus on your teaching.
Can you elaborate?
I used the same method and got the same solution, i^i=e^(-pi/2). But when I tried (i^i)^4 = (i^4)^i = 1^i = 1 != e^(-2*pi), why?
Exponents in the complex world are multivalued. So 1^i has many solutions, 1 being just one of them.
1= e^(i2npi), n€Z
1^i = (e^(i2npi))^i = e^(i2npi*i) = e^(-2npi)
For n=0 you get e^0 = 1.
For n=1 you get e^(-2pi).
For other n you get different solutions.
-It' far, far after midnight. What are you doing?
-I'm doing i to power i
I've seen this answered as i^i = (e^i*pi/2)^i = e^-pi/2 ~= 0.2. What I don't understand is that following this reasoning, it is also (e^5*i*pi/2)^i = e^-5*pi/2.
The answer is e^(-π/2 + 2πk), so your case is for k=-1.
i = exp[(90° + n•360°)•i]
i^i = 1/exp(90° + n•360°)
is a real and infinity solutions
In Euler formula, the angle must be expressed into radians not in degrees !
If you take a real number and raise it to i, you get e to the quantity of i times the natural log of the real number, which is equal to the cosine of the natural log of the real number plus i times the sine of the natural log of the real number.
This is how Euler’s formula works. It works for every complex number. Example four to the quantity of five plus three times i is equal to four to the five times e to the quantity of i times the natural log of four to the three. But here’s the question. If you take a complex number to a complex number, will the formula work? If it does work, here’s how you explain it with only math symbols: i^i = e^i ln i = cos(ln i) + i sin(ln i). But we can’t take the natural log of i. But oenr’s comment that is liked by the creator of this video says it’s real. And Josep Martí Carreras’s comment and SyberMath’s answer says it’s e^-pi/2. So the answer is absolutely e^pi/2.
please explain your point more clearly for a newb like me! I am a bit confused. Are you saying the author is wrong in this video!? Also How did he even get that initial equation in the beginning of the video. Where the heck does it come from?
What is e^h, where h is a quaternion(a + bi + cj + dk)?
🤯😲🤯
great video. Can you explain Cauchy's theorems? I have never understood Complex analysis because it was never explained to me in College. We just memorized the results and used them in examples. I would love someone to explain it because I have no idea how Cauchy came up with any of it. The man must have been an abstract genius.
so is this solution a result of bad math or this solution is ACTUALLY the solution and the graph of x^x would intersect the solution if that graph was to be drawn in complex plane too.? someone elaborate please
(-i)^(-i) resulta no mesmo número, results the same number.
But is e^(-pi/2) transcendental?
Yes
Very interesting video!! I like learning new things!! Thank you!!!!!❤❤❤❤
You are so welcome
1:14
Why?
I mean, why the number with e?
Why don't we say any other number for example 5^(ix) = cosx + isinx ?
e is a special number lne= 1
cos(i)=??? Imaginary or Real?
Euler's identity: e^(ix) = cos(x) + i.sin(x)
So: e^(-ix) = cos(-x) + i.sin(-x) = cos(x) - i.sin(x)
Adding them, we get: e^(ix) + e^(-ix) = 2.cos(x). So cos(x) = (e^(ix) + e^(-ix))/2.
When x = i, we have cos(i) = (e^(i.i) + e^(-i.i))/2 = (e^(-1) + e^(1))/2 = (e + 1/e)/2, which is a real number. Also equal to cosh(1) by definition.
Cos domain real number so ta i gote imagine number for this ama function i deiparibani
In quale riferimento ad esempio l ' imbattibilita' interplanetaria è determinabile? Che disastro !
e^i* pi/2 = i, So (e^i * pi/2)^i = e^(-pi/2) [ As i^2=-1] So i^i is Real and it is e^(-pi/2)
i would say. i too the power of i is just more i but not r.
You can't just multiple powers of i time i since they are complex numbers. That rule doesn't apply to complex numbers, even though you got lucky the answer came out correct. That rule only applies to real numbers. Please re-think is over and make the necessary correction to the steps. You are on the right track.
Correct 💯
are videos on the internet made by criminals who hide behind camera operators real?
"so, it's about a fifth" : M Parker.
I checked with a calculator and I got the same number, even more accurate.
Ничего не понимаю i^i также легко вычислить как например 4^4.
so its indeterminate if there's a k?
e^ipi = -1
sqrt(e^ipi) = sqrt(-1) = i
(e^ipi)^1/2 = i
e^(i^2pi/2) = i^i
This could have been solved in a few seconds!
Simply replace base i by e^i*pi/2 , raise it to power i then multiply the exponents to get -pi/2
Why such an elaborate video?
What is the point of imaginary numbers?
The set of Real and imaginary numbers put together is a complex number set. In other words all real and imaginary numbers are called complex numbers.
Is 3+4i an imaginary number? 🤔
@@SyberMath no it is a complex number. Any number of the form a+ib is a complex number in which a is the real part and b is the imaginary part. An imaginary number is something that cannot be represented on a number line.
i for an i make whole world blind
- MATHma Gandhi
*Jai ʂri Ram, 🕉️ 🚩*
Here i for an i gives us one-fifth of whatever we had.
This is the advanced class....
hallo auther. I wonder if (a*i)^i belongs to a Realnumber, provided that a belongs to R.
Hello! yes it does
i^i = e^-2pi•n x e^-pi/2
There are literally an infinite number of answers.
I have a challenge for you:
(7^x) + (4^x) = (13^x)
x = 0.7987
@@ulrichkaiser3794 How did you solve it!?
@@Peruvian_GianlucaPlayz By means of GeoGebra plotting 7^x + 4^x -13^x
and looking up the ( x / 0 ) position ... ;-)
@@ulrichkaiser3794 sorry, i don’t understand
not real is not the same as imaginairy.
as classification. for example.
the number 10 is a part of the number 100
axioma's
1.real is not imaginairy
2.reality alowes imagination.
therefor.
hypothesis:
1.imagination is a part of reality bit not the whole reality, true or false ?
so of 10 is a part of a bigger number. 100
things you dont imagine and are not part of reality can also excist.
then
negative 10 is not a 100
but 10 too the power of 10 is.
so the question you ask is.
does your imagination covers the whole reality ?
or.
do you have an accurate view on reality
or
do you intrepretate reality well, without observing it ?
is that the line of your real question ?
or are you missing something. like is the language too abstract too describe reality.
(the answer on the last one is, yes, the language is merely a symbolicle representation of the real thing but sufficient enought to work with. )
or are you just saying hi...
what is the moral of the story here? I watched the whole video but the guy never makes a point at the end as to why he did all of this!
make a video why 1 to the power Infinity is not 1
your understanding of math should be measured by how far you can make it through this video before you give up trying to understand it and just watch to see what the answer is
So... i*ln i is -P/2?
What will be the answer of i^2i
Anybody???
Yes
Seems this is a required problem for all math sites
imaginairy stated as conciouss descision.
This is unreal, man!
Really!!!
It's 1/e not e^-pi/2
Что то я не очень понял, но очень интересно! 😆
Very cool video!
Thank you very much!
My guess: (e^ipi)=-1 so (e^(ipi/2))=i so i^i= e^(i^2pi/2)=e^(-pi/2) so using a calculator .2078…
arg 1 = 0, arg -1 = pi, arg i = pi/2, arg -i = -pi/2
arg function is the angle in radians.
In angles, arg 1 = 0, arg -1 = 180, arg i = 90, arg -i = 270 or -90
It’s only angles. Not radians.
"I'm gonna go get the papers, get the papers."
Imaging, imagining is quite real.
😲😀
Why in the world did you bring Euler's Identity into it? Completely unnecessary. Why didn't you write i = exp((i*Pi)/2) and then use your rules of exponents. Would have yielded the same result in less than 1 line.
Where does that come from?
(e^(i*pi/2))^i=e^(-pi/2)...real
When you show this to a teacher, that person will faint.
Be careful when presenting graphs of solutions without proper identification of your graphs. You went from presenting the solution of i=cosx+isinx on the complex plane to presenting a solution of i^i=e^(-pi/2) on the real axis. Teaching complex numbers can be a bear ...
very interesting ! respect from china
Thank you! 🤗
i^i=roughly 1/5. Damn...
exp(-pi/2)
Imaginary indicates a Tachyon enigma, Thanks, so I can become Etc., however, how do You pls. set that 'e' be? Thanks!
nigga balls
愛の愛情は実数
Yes it is.
Great !
Let
i^i = r[cos(a)+i*sin(a)] ---- (1)
(i^i)^2=r^2[cos(2a)+i*sina(2a)]
i^(i^2)=
i^-1=
1/i=
i/(i*i)=
i/(-1)=
-i=
Then by comparing the real part and imaginary part, we can have the r and a.
Sub r and a into (1), then problem solved. answer. 😊😊😊😊😊😊
How to prove i^i>1/5 without using calculator
glad i knew about eulers formula and could instantly arrive at the answer
#bro face reveal
Nice!
Thanks!
((-1)^0.5)^(-1^0.5) = ±1^0.25 ...
0:49 He fails a little
пишут так что бы сбить с толку!
2:20
экспоненциальная запись это е в степени i * (фи), а если вместо фи используется х то и на рисунке должно быть х, но на рисунке пишут не х, не фи... а большую Тета!
Вот зачем такие запутки?
Написал х и на рисунке пиши х!
А если не х то изначально пиши Тета и не морочь голову!
2:50
А объяснение синуса и косинуса через r, а потом в итоге выписано классическое определение синуса и косинуса (отношение определенного катета к гипотенузе)..... Жесть! Все сделано через левое плечо и правое колено!
Жесть! синус есть синус, а косинус есть косинус! На кой тут r применили.... (я знаю что такое r но тут оно не к месту!)
3.00
а зачем писать нормальную форму комплексного числа для i и еще уточнять что мы имеем r=1 ? Вот на кой это?
3:30
ну да-да-да, 30 сек рассуждаем как комплексное число i будет отображаться на комплексной плоскости.
Простите но это очевидно! (исходя из нормальной формы записи числа i)
3:50
записали i в тригоном.форме... а на 2:20 почему это нельзя было сделать?
короче... до сих пор это все теория....
ладно... пусть так!
4:20
это можно было написать тогда когда экспоненциальную форму описывали т.е. 2 минуты назад....
(ок не ворчу......)
4:45
ну да.... только это одно действие и есть решение, а все остальное это теория..... которую должны знать те кто решают задачи с комплексными числами! (пол ролика теории.... нуладно! если автор так видит...)
5:20
вспоминает про период в два Пи.... мда.... печально! Вообще то это изначальное определение! И опускание этого факта в самом начале - это ошибка!
короче!
Баба Яга против! (ну не кошерно это!)
real 🤣: i=exp(i*pi/2) so i^i = exp(-pi/2)
Real ?
Won.
🤍 amazing
Thank you
i^i = 1/8th root of -1
reasoning?
答え i^i= e ^-Πi
Wolfram say it's real: e^(-π/2)
Eu pensei nisso na escola e todos disseram que sou idiota😮
if i = -1 × real.
if imaginairy is a negative real.
there are a lot of "notreal" but they are not all the opposite of real. yhe words come from latin or greek.
1real
2.anti real
3.contra real
4.pseudo real
5.quasi real
6.beyond real
7.before real
8. areal
9.non real
10.not real
11.advanced real
12.not real yet
13.was real once but not know
14.is real for you but not for me
15.phantasy
16.imaginairy
17.like a dream
18.surreal (more then real)
something like that. imaginairy is not that black and white or diabolic as presumpted.
like if real state too imaginairy then
if turning left state too going up... as ?
instead of simple turning right.
it is possible that imaginairy is more .. then reality.. although it is reality that makes imagination possible.
if you ask me... everything is real. found out have a very efficient minimal perceptual view on the matter. senses are body specific. so pretty sure reality is more then can be perceived.
1^10=1
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1
If i=e^i.pi/2 then (i)^i= (e^i.pi/2)^i
Here folowing you will making a logic syntax error:
You get power i in processin into e^i.pi/2 and resulting
i^i= e^[(i^2).pi/2]
You can not go like that!
wow nice
Thanks