Russia Math Olympiad | How To Solve For All 4 Imaginary Roots In This Nice Exponential Equation.
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- เผยแพร่เมื่อ 18 เม.ย. 2024
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Putting a^4 = - 4 in the form of Euler’s identity: a^4 = 4*e^i*(Pi + 2*n*Pi), therefore a = sqrt(2) * e^i*(Pi/4 + n*Pi/2). Visually, the 4 roots are located on a circle of radius sqrt(2), evenly spaced with angles Pi/4, 3Pi/4, 5Pi/4, and 7Pi/4. In real and imaginary coordinates, that’s +/- 1 +/- i. Most of these problems are easier to solve by switching to polar coordinates using Euler’s equation, finding the obvious solutions with hardly any calculation, and then switch back to Cartesian coordinates (real and imaginary coordinates).
Superb!
😢😢😢😢😢😢
Yes it is indeed that simple, but this was really a question in a Russian Math Olympics?
solving this problem is not for the calculus level. that's why he solved it using algebra 2.
@@fadz5210
Math Olympiads are competitive events where speed is crucial. The method described in the video is the method the loser would use…
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Enjoyable tutorial, I was able to follow and understand every step. I just LOVE ALGEBRA! It is such a relaxing pastime for me!
Wow!!!
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Beautiful display of math! Thank you for sharing!
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Use Euler's equation to quickly solve any equation a^m=-b where m is a positive integer and b≥0 is real.
The trick is to first write a^m=b*(-1)=b*e^(iπ+i2jπ), j=0,1,...,m-1
Then a_j=b^(1/m)*e^(iπ/m+i2jπ/m)=b^(1/m)*[cos(π/m+2jπ/m)+i*sin((π/m+2jπ/m)], j=0,1,...,m-1
where a_j, j=0,1,...,m-1 represents the m mth roots of (-b)
Wow!! I will try applying this method although I have used the Euler's equation in solving many math problems on this channel. Thanks for the suggestion sir.
And i^i = i. ? 😮😮
Można dużo prościej !!! Wyjść od (a^2 + 2i)(a^2 - 2i)=0 szybko i prosto, mniej pisaniny! I o to chodzi w matematyce.
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Very good .,explanation. Thank you.
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Long winding route. Ultimately it's root of -1this can b done in 3 steps. No need to add and make it look complicated.
Nicely explained. ❤❤❤
Thanks a million for watching and at the same time dropping this wonderful comment sir.
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Beautifully solved man.
Thanks a bunch for watching sir.
Thanks for the comment also.
Much love sir
Nice approach master
Thanks a bunch sir
Clear explanation thanks
You are most welcome sir.
Thanks for watching and commenting boss.
Much respect!!!
Great!!
Thanks a million sir
Much love sir.
Here is another way to solve this equation and I believe this might be little easier
1. a^4 = -4
2. (a^2)^2 = 2^2 * (-1) (Case A) or (-a^2)^2 = 2^2 * (-1) (case B)
3. Let us solve Case A first, where (a^2)^2 = 2^2 * (-1)
4. Taking square root of both side a^2 = 2i
5. Taking square root of both side a = ✓2*✓i
6. Square root of i = 1/✓2 + i/✓2 and -(1/✓2 + i/✓2)
7. So a = ✓2*(1/✓2 + i/✓2) and a= ✓2*-(1/✓2 + i/✓2)
8. So a = (1+i) or a = -(1+i)
9. Now take the Case B, where (-a^2)^2 = 2^2 * (-1)
10. (-a^2)^2 = 2^2 * (-1)
11. Taking square root of both side -a^2 = 2i
12. a^2*i^2 = 2i
13. Taking square root of both side a * i = ✓2*✓i
14. Or a = ✓2/✓i
15. Square root of i = 1/✓2 + i/✓2 and -(1/✓2 + i/✓2)
16. So a = ✓2/(1/✓2 + i/✓2) and a = ✓2/-(1/✓2 + i/✓2)
17. So a = (1-i) or a =-(1-i)
18. Four values of a = (1+i) or a = -(1+i) or a = (1-i) or a =-(1-i)
Wow! this pretty cool.
Bravo!!!
You the best sir.
Maximum respect sir 🙏🙏🙏
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Thank you very much. Even 12:58 my 13 year old grand daughter is fascinated by your clear and comprehensible explanation of your math problems. Keep up the stupendous work. Thank you from the Negev in the land of Juda.
Nice work my brother.
Keep it up
Thanks brotherly.
Amen!!! 🙏🙏🙏
Thanks
Very nice process of derivation by applying a few very useful techniques. My approach is just doing this. a^4 -> (a^2)^2, -4 -> (2i)^2 and -4 -> ((-2)*i)^2, so a1^2 = +/-sqrt(2i) and a2^2 = +/-sqrt(-2i). 2i -> (1+i)^2 and -2i -> (1-i)^2, so a1=+/-(1+i) and a2=+/-(1-i). Therefore the final solution is a1=1+i, a2=1-i,a3=-1+i, and a4=-1-i.
Very good!
Ajab khan khattak.I solve it as:a^4 =- 4
(a^2)^2 =-(2)^2
a^2 =-(2)
a=-2under root
Wow!!! This approach is not complete sir.
Good work, maestro!
Thank you .good luck
Thank you! You too!
Your the best sir
Great
Thanks very much sir.
We love you for watching and commenting sir.
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ممتع ،، من العراق
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fine! Thank you!
Smiles...👍👍
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Very nice derivation and explanation!
Might I suggest you show all four answers on the real/imaginary plane so the students can see how symmetrical the graphed points are?
Ok sir, thanks for this suggestion.
Noted sir.
Thanks 👍
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I was trying to do it directly by stating the difference of two squares using i. (a^2+2i)(a^2-2i)=0
But now I end up needing to square root the imaginary, meaning I would have to set it up to a+bi. Your creative way turns out simpler, although that too had a lot of steps
Hahaha....I tried that approach also but led me to a more complex stage hence I resolve to this method.
Thanks a million for the attempt sir.
You the best sir.
Nice video
Thanks a million for encouraging Onlinemaths TV by watching and commenting all the time sir.
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Very nice
Thanks for watching and at the same time leaving a comment behind sir.
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Why did I get different complex roots? I got sqrt(2i), -sqrt(2i), sqrt(-2i) and -sqrt(-2i). When you plug these roots in the equation it is true. Could someone explain?
thats the roots i found! Could someone explain?
That's the root bro
You are also very correct.
That is also a solution that satisfy the original equation sir.
You the best sir.
Меня восхищает, когда квадратное уравнение a² + 2a + 2 = 0 решается с помощью формулы для корней. Это же гениально. А если переписать это в виде (a + 1)² = -1; не станет ли это проще, понятнее и ну как-то ... разумнее, что-ли?
Надо сказать, что я когда-то "сел" (опростоволосился) на первом курсе. Нужно было взять интеграл от 1/(x⁴ + 1), а я не догадался, что знаменатель раскладывается в произведение двух многочленов второй степени. Справедливости ради стоит сказать, что до того момента я никогда не слыхал об основной теореме алгебры. На самом деле любой многочлен четвертой степени с действительными коэффициентами может быть разложен таким образом, так как комплексные корни могут существовать только парами сопряженных. В точности как в этой задачке. Детям сейчас это рассказывают, а вот нам не рассказывали, от слова совсем.
Wow!!! that is a wonderful approach from you sir, it is very short, explicit and short.
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Much and maximum respect sir.
Very nice explanation.
Lots of love and respect from India 🇮🇳.
Smiles, thanks a million sir.
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@@onlineMathsTV Thank You So Much👏
Le faltó comprobar que a^4=-4. Cualquier profesor de matemáticas sabe que hay que comprobar las soluciones con el único propósito de verificar que la respuesta sea cierta
I check all roots and they all satisfy the original problem.
If you want me to produce a video just to verify the roots then drop it in the comments section and I will make a video for the verification sir.
Thanks a million.
👍👍👍
Thanks a bunch sir.
-4=4*ехр(i*(pi+2*pi*k)), k- integer, pi=3,1415..., i -imaginary unit.
ok
Классика жанра:
Четвертая степень => четыре корня!
very complicated solution. could be much easier with the trigonometric representation of complex numbers.
nice job ... but a few suggestions
first: would be clearer to avoid "a,b and c" in the original formula -- use z^4 = -1
second: you already have results 1,2,3,4 ... why rename them as results 3,4,1,2 ???
third: you gain nothing by referring to one of the results as -(1 + i) === -1 - i
(negation has precedence over subtraction)
Thanks for this nice observation sir.
Your comment is well noted sir.
a = e^(0.25ln(4) + 0.25ipi + 0.5ipin), n is a integer
Ok sir
Понравилось.
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i ❤ Mathematics
Really? Then you are my type without mincing words.
Am glad to meet with you sir.
Maximum respect sir.
As a math lover, pls your criticism are welcome , hahahhahaha.
Решаем уравнение, а^4=4, а затем находим корни на окружности с радиусом корень(2).
ok, I will check that approach out
Простое уравнение. Решение очевидно.
Sure, the answer is obvious depending on your level of math prowess.
Theorem. The solutions of the equation xⁿ=a, where a is a negative real number, are
xᵢ = ⁿ√∣a∣ ⋅ ԑᵢ (i = 0, 1, ..., n - 1),
where ԑᵢ are the nth roots of -1.
Noted sir.
Easy
Sure!!!
Sure!!!
If you stop to think about it, the roots of a^4 = -4 are rather obvious. Whatever a is, it has to be a divisor of -4. 2 and -2 are real divisors, but they won't do. On the other hand, (1 + i ) is a divisor of 2 and thus -4. One can easily check that is is a root. The rest flows from that.
No, they are *not* "rather obvious." *Learn* the proper contexts of when to use that phrase.
i hate the man who makes question. 🤣
but my man can solve this question easly
Hahahahaha... 🤣🤣🤣 funny comment.
Problem creators are really geniuses and we should love and respect them for their creative prowess because they make the whole world a beautiful place to dwell sir.
Thanks my good friend for the acknowledgement sir.
-1-i
хорошо, а куда делись корни a=+- root (+-2i) ??
ok, but where is a solution a=+- root (+-2i) ??
для начала неплохо было бы задать себе вопрос, что такое корень из i
тут все просто. Корень из i - такая величина, которая при возведении в квадрат имеет значение i.
@@Alexey_Alex1
a=(1+i)^4 0r a=(1-i)^4
🇩🇴🇺🇸🇩🇴🇺🇸
really need this comp?
Yes sir
No, it is too too complicated. There is much more easy way.
я из росии, но подача мне понятна
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Thanks a million sir
i*sqrt (2) 😉
How?
-1 = a
No correct sir.
Kindly watch the video from the beginning to the end without skipping any parts for proper understanding sir.
Thanks sir.
Put isquare is -1 then it will be 2i , think, what about u
Sorry, I do not get your point here sir.
Kindly rephrase your question or point.
Thanks sir.
Method 2?
Alright sir
Why not write the equation as a^4-4(i^2)=0.
Will that give us the needed need four roots?
I will check that out at my leisure time sir.
Thanks for this wonderful suggestion.
@@onlineMathsTV I worked it out and got four roots.
a=1+i ou a=-1-i
Bravo!!!
You are the master here boss.
Maximum respect 🙏🙏🙏
a=a or a=/a ???!!!
Not too clear with this sir.
- a
We are looking at the numerical value sir
We are looking at the numerical value sir
ok
And i^i = i. ?
i*i*i=-i
i*i=-1
@@user-jr6ue7rk9p pow( i, i ) ?
That statement is mathematically correct sir.
Very correct,
Bravo!!!
Bravo!!!
Hmmm
This math is somehow
Hahaha 😂😅😂😅
Smiles....how sir?
Just ask a question where you are not clear and I will explain sir.
,(√2√i)^4=-4
لماذا كل هذه الزوبعة ، كان يمكن الحصول على الحل بطريقة بسيطة جدا فور الاطلاع على المعادلة.
Really?
So what is the shortest or best and unique approach to this math challenge sir?
Kindly drop your 1second approach to this math problem so that we all at Onlinemaths TV can learn from you sir.
We are open to learn from gurus like you boss...🙏🙏
Thanking you in advance sir.
Hay un error en el enunciado porque ésto parte de un error ,de acuerdo a la teoría de exponentes todo número elevado a un exponente par ya sea el número positivo o negativo ,siempre se tendrá como resultado un número positivo
Really?
عمال الوكالة الرابعة بن طلحة ومالحقة المديرية العامة بالمحمدية وكذا خلية الأمن الداخلي إضافة داءرة الصيانة وكذا مصالح داءرة المحاسبة تورطو في اتلاف مقررات عدالة وقراري ترقية وكذا الإعداد لحوادث مع فرقة الصيانة ببن طلحة وحي الفين واربعة مسكن وحوش ميهوب عن طريق عناصر تمنية متخفية تشغل مناصب في الديوان كما اعدو لحوادث خطيرة كالحراءق المنزلية مع مؤسسات كالداءرة و البلدية بتواطؤ مصالح الأمن شرطة جيش ودرك وامن بمساندة دول اجنبية كتركيا وفرنسا وانجلترا واوكد الدور الذي لعبتهمضباط واطارات امنية حضرت ابن طلحة لاعطاء إشارات ان الدولة سترعى المجرمين ضد الشاب اوباغة طاهر كما أذكر دور سكان الاحياء كي الف وثمتنية وسبعين مسكن في فبركة الحوادث و القيام بالضغط بتواطؤ عمال الوكالة
Hahahaha...😂🤣😂🤣.
Is this comment for this video tutorial? I doubt!!!
knesolvh tebedda kalbeqahtebedadadadadadas...
I couldn't translate your comment but thanks so much for watching and dropping this comment.
I couldn't translate your comment but thanks so much for watching and dropping this comment.
a^4 > or =0 => a^4= -4 wrong 😤
Hahahaha....🤣😂🤣😂🤣
This is to long
so sorry about that, I will adjust the timing in subsequent video sir.
Thanks a million for this observation boss.
are you teacher or professor ? i don't know who you are , but you have a math ( unreal ) . Every body knows a2 a4 a6 ….> 0 with every value of a except a = 0 , a2 a4 a6 …. = 0 , in my opinion i don't think a4 = - 4 ( is it real or a riddle ) . but you have answer . why don't you check again with your answer ( with value you find ) . for more basic math knowledge : √a >=0 , icase a
But imaginary numbers (with negative squares) are as realistic as negative numbers are - they are "perpendicular" numbers, just like negatives are "backward" numbers. Both of them have similar angle representation:
+1 = 1∠0° (to the right)
−1 = 1∠180° (to the left)
+𝒊 = 1∠90° (up)
−𝒊 = 1∠270° (down)
And you can use this representation to solve equations and check solutions.
Many thanks for your detailed explanation sir.
Maximum respect to you sir 🙏🙏🙏
Thanks for watching and dropping this wonderful observation of yours sir.
You have said well, but try as much as possible to make some research works before reaching a concrete conclusion.
もっと簡単にできる方法はないか
try to solve a^4=-4 with t=a^2. I don't sure if it's easier but it's gonna be fun
(now t^2=-4, you have tofind all 't' for this one equal and backward replace 't' to a^2)
I will try if the solution to this problem can be figured out in an easier way and I will get back to you as soon as I figure that out.
Just give me a soft reminder should it escape my mind to check.
Thanks a bunch sir 🙏🙏🙏
And if it were ⁴√x = −4 instead, would it have real or complex solutions then?
Nice question sir.
I will a video on this math problem immediately and see the nature of roots it will give sir.
Kindly watch out for the video on this channel.
Thanks a million sir.
@@onlineMathsTV Please let me know here in the comments when the video is out, so that I don't accidently miss it.
It is a very good mathematically deep question indeed which is not that easy to give a definite answer to.
Thanks