Imaginary numbers are so cool because even if they are called "imaginary" and everyone calls them "made up shit", they actually make so much sense and thats so fucking cool
Assumption: For all a,b,c holds: (a^b)^c = a^(b*c) => 1 = e^(2*π*i) = e^(2*π*i)^(2*π*i) = e^((-1)*4*π^2) < 1 => There is an a,b,c in the Complex Numbers such that: (a^b)^c ≠ a^(b*c). This exponential rule does not necessarily apply for complex numbers, so if you want to use it, you must prove that it works in your specific case. If you want to use complex analysis in general, you have to be careful. Derivatives, functions (like 'log(x)'), Integrals and geometry behave in general a bit differently on the complex numbers. The journey of learning mathematics is full of traps and mistakes(I myself made many of those), which makes one smarter if seen correctly. To see those correctly, one must try to filter out the hidden conditions behind the things used, which can be difficult. From there on its just logic, which itself can sometimes be a bit unapparend. I hope this will help you on your way. Good Luck :)
@@expmine Hi brother! Thank you for your comment. I love the fact you pointed that out, it turns out (a^b)^c = a^(bc) is very nuanced and actually has many rules to it. When replacing b and c with real numbers, the equation always holds true. However, as you said, for complex numbers it isn’t always true. For the purposes of this video, I used the principle branch of i, which is e^(ipi/2) not exploring alternative branches with + 2npi. Explaining and proving why the exponential rule doesn’t work for alternative branches is a bit more in depth, and I can do a video on it if you would like! However if we replace b and c with the principle definitions of imaginary numbers (quarternions and the set H is excluded from this rule for a separate reason), the exponential rule holds true. Hope that helps!
@@amaarculus Thanks for the answer. This helps very much, because I forgot, that the rules of exponentials are different on the principle branch, since complex analysis and the algebraic properties are not things I primerly focused on. If one uses the rule (a^b)^c = a^(b*c) for complex numbers, then one must be very careful, since the set on which it works depends on a. (See edit for more Information) Edit: The proof, that (e^a)^b = e^(a*b) for a and b on the princible branch is not that difficult, if one uses that the logarithim is unique on the princible branch and shows that it does not work on the other Branches by using a contradiction (That is what I have done on my whiteboard). But if one wants to prove (a^b)^c = a^(b*c), then the period on which this works will change for many a (since I just looked at real numbers. It may or may not affect most or all) , hence one would not use the typical principle branch. Did/Do you study math by any chance?
So if I dream about my dream, dream no. of times, it becomes real right? But I shouldn't dream about me dreaming about my dreams cause that'll make it just imagination again, right???? Or am I trippin.
Dang, that's wild. I guess every even exponent of i is a real number due to every even exponent of i being a real number as well. This is very interesting for a nerd like myself who also hasn't worked a day in his life on any imaginary numbers
The square root function, indicated by the radical sign, is defined (as you should know if you use any function) from the positive reals to the positive reals. This means that by definition, any input that is not a positive real has an undefined image by this function or map. You cannot algebraically manipulate this to say that i = sqrt(-1) (I am a student in physics at the ENS Lyon if you want my qualifications). Not only is it not "technically" correct, it's just plain incorrect. You can however, define (or if you're lazy, assume that it is already defined) a function or map that takes all reals (or even all complex numbers) as antecedents and returns a complex (which as you know will also be real if, for example, the antecedent is a positive real). But you cannot use the radical sign as it is an elementary function that is defined as the principal square root from R+ to R+. This is exactly what happens with the use of complex branches of the log function: even if it is previously defined to be an extension of a known function (here log) to larger sets, it's messy and really not very useful for anything practical. So the correct (albeit messy) way to say it is "should complex_sqrt be the function from C to C that returns the principal square root of the antecedent (up to you how you want to calculate it since you're the one who defined it), then we can algebraically deduce from i^2 = -1 (proper definition of i, as well as "one of the solutions to the equation x^2 + 1 = 0"), that complex_sqrt(-1) = 1" because we have a function that allows us to do so.
@@JaxEntersEvasion Doing all that with a polynomial space seems to be a bit too complicated for my taste. I personally prefer the definition by isomorphism of R^2, in which the (complex) squareroots of -1 are defined by the multiplication formalism for complex numbers. With this, one can define the direct sum of span{1} and span{i}, where i is one of the squareroots of -1, since its properties can be defined through the isomorphisim, which itself exists since the direct sum is finite dimensional. With this, one has a space in which the squareroot of -1 exists and it is called i, as well as one has the typical formalism with i = +sqrt(-1). Also does this definition give rise to a simple definition of complex differentiability and many other things (This is how I learned the complex numbers).
![30 ชั่วโมงคนเดียว บนรถไฟชั้น 3 โซเวียต [Around The World Ep. 2]](http://i.ytimg.com/vi/MlE9boAVqDA/mqdefault.jpg)
Imaginary numbers are so cool because even if they are called "imaginary" and everyone calls them "made up shit", they actually make so much sense and thats so fucking cool
Assumption: For all a,b,c holds: (a^b)^c = a^(b*c) => 1 = e^(2*π*i) = e^(2*π*i)^(2*π*i) = e^((-1)*4*π^2) < 1 => There is an a,b,c in the Complex Numbers such that: (a^b)^c ≠ a^(b*c).
This exponential rule does not necessarily apply for complex numbers, so if you want to use it, you must prove that it works in your specific case. If you want to use complex analysis in general, you have to be careful. Derivatives, functions (like 'log(x)'), Integrals and geometry behave in general a bit differently on the complex numbers.
The journey of learning mathematics is full of traps and mistakes(I myself made many of those), which makes one smarter if seen correctly. To see those correctly, one must try to filter out the hidden conditions behind the things used, which can be difficult. From there on its just logic, which itself can sometimes be a bit unapparend. I hope this will help you on your way. Good Luck :)
@@expmine Hi brother! Thank you for your comment. I love the fact you pointed that out, it turns out (a^b)^c = a^(bc) is very nuanced and actually has many rules to it. When replacing b and c with real numbers, the equation always holds true. However, as you said, for complex numbers it isn’t always true. For the purposes of this video, I used the principle branch of i, which is e^(ipi/2) not exploring alternative branches with + 2npi. Explaining and proving why the exponential rule doesn’t work for alternative branches is a bit more in depth, and I can do a video on it if you would like! However if we replace b and c with the principle definitions of imaginary numbers (quarternions and the set H is excluded from this rule for a separate reason), the exponential rule holds true. Hope that helps!
@@amaarculus Thanks for the answer. This helps very much, because I forgot, that the rules of exponentials are different on the principle branch, since complex analysis and the algebraic properties are not things I primerly focused on.
If one uses the rule (a^b)^c = a^(b*c) for complex numbers, then one must be very careful, since the set on which it works depends on a. (See edit for more Information)
Edit: The proof, that (e^a)^b = e^(a*b) for a and b on the princible branch is not that difficult, if one uses that the logarithim is unique on the princible branch and shows that it does not work on the other Branches by using a contradiction (That is what I have done on my whiteboard). But if one wants to prove (a^b)^c = a^(b*c), then the period on which this works will change for many a (since I just looked at real numbers. It may or may not affect most or all) , hence one would not use the typical principle branch.
Did/Do you study math by any chance?
Very nice explanation!
Can we for i^(i^i) expand it as (i^i)^i = (exp(-π/2))^i = exp(-iπ/2) = -i?
Good explanation.
Hilarious, a number that can be real or imaginary depending on the exponent 🙀
So if I dream about my dream, dream no. of times, it becomes real right? But I shouldn't dream about me dreaming about my dreams cause that'll make it just imagination again, right???? Or am I trippin.
You’re right
Dang, that's wild. I guess every even exponent of i is a real number due to every even exponent of i being a real number as well. This is very interesting for a nerd like myself who also hasn't worked a day in his life on any imaginary numbers
i is not the square root of negative one. Please stop propagating the american lack of mathematic formality.
If i^2 is defined as -1, how is the square root of -1 not i?
whats i then
If i is not equal to the square root of negative one, my asian school failed me
The square root function, indicated by the radical sign, is defined (as you should know if you use any function) from the positive reals to the positive reals. This means that by definition, any input that is not a positive real has an undefined image by this function or map.
You cannot algebraically manipulate this to say that i = sqrt(-1) (I am a student in physics at the ENS Lyon if you want my qualifications). Not only is it not "technically" correct, it's just plain incorrect. You can however, define (or if you're lazy, assume that it is already defined) a function or map that takes all reals (or even all complex numbers) as antecedents and returns a complex (which as you know will also be real if, for example, the antecedent is a positive real). But you cannot use the radical sign as it is an elementary function that is defined as the principal square root from R+ to R+.
This is exactly what happens with the use of complex branches of the log function: even if it is previously defined to be an extension of a known function (here log) to larger sets, it's messy and really not very useful for anything practical.
So the correct (albeit messy) way to say it is "should complex_sqrt be the function from C to C that returns the principal square root of the antecedent (up to you how you want to calculate it since you're the one who defined it), then we can algebraically deduce from i^2 = -1 (proper definition of i, as well as "one of the solutions to the equation x^2 + 1 = 0"), that complex_sqrt(-1) = 1" because we have a function that allows us to do so.
@@JaxEntersEvasion Doing all that with a polynomial space seems to be a bit too complicated for my taste. I personally prefer the definition by isomorphism of R^2, in which the (complex) squareroots of -1 are defined by the multiplication formalism for complex numbers. With this, one can define the direct sum of span{1} and span{i}, where i is one of the squareroots of -1, since its properties can be defined through the isomorphisim, which itself exists since the direct sum is finite dimensional. With this, one has a space in which the squareroot of -1 exists and it is called i, as well as one has the typical formalism with i = +sqrt(-1). Also does this definition give rise to a simple definition of complex differentiability and many other things (This is how I learned the complex numbers).