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).

@vicentegomes7270

7 күн бұрын

Superb!

@kitsiongphang3989

7 күн бұрын

😢😢😢😢😢😢

@markslowhand4214

6 күн бұрын

Yes it is indeed that simple, but this was really a question in a Russian Math Olympics?

@fadz5210

4 күн бұрын

solving this problem is not for the calculus level. that's why he solved it using algebra 2.

@alfredofettuccine9425

4 күн бұрын

@@fadz5210 Math Olympiads are competitive events where speed is crucial. The method described in the video is the method the loser would use…

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)

@onlineMathsTV

11 күн бұрын

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.

@josedivinodossantos1634

8 күн бұрын

And i^i = i. ? 😮😮

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.

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

Nice work my brother. Keep it up

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)

Good work, maestro!

Very nice

Beautiful display of math! Thank you for sharing!

@onlineMathsTV

11 күн бұрын

Hahahaha....thanks a tone the boss and thank you for watching our contents, and leaving a mind blowing encouraging comment sir. Maximum respect sir. We at Onlinemaths TV love you dearly ❤❤

Dear, Warm greetings from Bohnsdorf, Germany. I wanted to take a moment to express my sincere appreciation for the incredibly informative video you recently posted. Your content is not only highly engaging but also remarkably clear and understandable. I've taken the liberty to subscribe to your channel, and I eagerly anticipate the release of your future videos. Your dedication to providing valuable content does not go unnoticed, and I'm grateful to be part of your audience. Thank you once again for your efforts in creating such enriching content. Best regards

Very good .,explanation. Thank you.

@onlineMathsTV

11 күн бұрын

Thanks sir for watching and encouraging us in your comment. Much love from Onlinemaths TV ❤❤

@amudangopal

10 күн бұрын

Long winding route. Ultimately it's root of -1this can b done in 3 steps. No need to add and make it look complicated.

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.

Thanks 👍

Można dużo prościej !!! Wyjść od (a^2 + 2i)(a^2 - 2i)=0 szybko i prosto, mniej pisaniny! I o to chodzi w matematyce.

@onlineMathsTV

11 күн бұрын

Ok sir, noted for subsequent videos. Thanks for this contribution sir. Much love ❤❤

a = square root 2*i

i hate the man who makes question. 🤣 but my man can solve this question easly

Nicely explained. ❤❤❤

@onlineMathsTV

11 күн бұрын

Thanks a million for watching and at the same time dropping this wonderful comment sir. We love you from the depth of our hearts...❤❤

-4=4*ехр(i*(pi+2*pi*k)), k- integer, pi=3,1415..., i -imaginary unit.

Beautifully solved man.

@onlineMathsTV

11 күн бұрын

Thanks a bunch for watching sir. Thanks for the comment also. Much love sir

👍👍👍

very complicated solution. could be much easier with the trigonometric representation of complex numbers.

What about a = +/- sqrt(2i) ?

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.

@onlineMathsTV

4 күн бұрын

Very good!

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?

@onlineMathsTV

11 күн бұрын

Ok sir, thanks for this suggestion. Noted sir.

Ajab khan khattak.I solve it as:a^4 =- 4 (a^2)^2 =-(2)^2 a^2 =-(2) a=-2under root

@onlineMathsTV

11 күн бұрын

Wow!!! This approach is not complete sir.

Понравилось.

Решаем уравнение, а^4=4, а затем находим корни на окружности с радиусом корень(2).

i ❤ Mathematics

@onlineMathsTV

11 күн бұрын

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.

Простое уравнение. Решение очевидно.

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

Very nice explanation. Lots of love and respect from India 🇮🇳.

@onlineMathsTV

11 күн бұрын

Smiles, thanks a million sir. Maximum love from all of us at Onlinemaths TV to you and your wonderful family sir...❤❤

@gurunirankari4776

5 күн бұрын

@@onlineMathsTV Thank You So Much👏

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)

@onlineMathsTV

11 күн бұрын

Thanks for this nice observation sir. Your comment is well noted sir.

i*sqrt (2) 😉

a=1+i ou a=-1-i

-1-i

🇩🇴🇺🇸🇩🇴🇺🇸

хорошо, а куда делись корни a=+- root (+-2i) ?? ok, but where is a solution a=+- root (+-2i) ??

@alexeyalex8482

8 күн бұрын

для начала неплохо было бы задать себе вопрос, что такое корень из i

@dimitrykhavzhu5697

Күн бұрын

тут все просто. Корень из i - такая величина, которая при возведении в квадрат имеет значение i. @@alexeyalex8482

Why not write the equation as a^4-4(i^2)=0.

@onlineMathsTV

11 күн бұрын

Will that give us the needed need four roots? I will check that out at my leisure time sir. Thanks for this wonderful suggestion.

@leetrask6042

11 күн бұрын

@@onlineMathsTV I worked it out and got four roots.

And i^i = i. ?

@user-jr6ue7rk9p

8 күн бұрын

i*i*i=-i i*i=-1

@josedivinodossantos1634

8 күн бұрын

@@user-jr6ue7rk9p pow( i, i ) ?

Hmmm This math is somehow

@opulence3222

13 күн бұрын

Hahaha 😂😅😂😅

@onlineMathsTV

11 күн бұрын

Smiles....how sir? Just ask a question where you are not clear and I will explain sir.

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

@allozovsky

8 күн бұрын

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.

This is to long

もっと簡単にできる方法はないか

@nikolyangelo

Күн бұрын

try to solve a^4=-4 with t=a^2. I don't sure if it's easier but it's gonna be fun

@nikolyangelo

Күн бұрын

(now t^2=-4, you have tofind all 't' for this one equal and backward replace 't' to a^2)

And if it were ⁴√x = −4 instead, would it have real or complex solutions then?