How To Find The Square Root Of Iota, ί. √ί=? | square root of iota, ί.

How To Find The Square Root Of Iota, ί, is a video tutorial designed to give you a clearer picture into the world of imaginary numbers in mathematics.
In this video, Jakes x-rayed the best simplification approach to this proof leaving no stone unturned.
Watch from the beginning to the end for full understanding leave a comment in the comments section.
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Пікірлер: 96

  • @nagarajahshiremagalore226
    @nagarajahshiremagalore226Ай бұрын

    Excellent presentation of steps to find square-foot of Iota. Thanks for uploading this video coaching.

  • @kee1zhang769
    @kee1zhang76911 ай бұрын

    We have a^2=b^2 implies that a=b or a=-b. But a=-b does not satisfy the equation 2ab=1, so we reject a=-b. There are 2 answers: namely (1+i)/sqrt(2) or -(1+i)/sqrt(2) . The first answer is called the principal square root of i, which can be found easily by Euler's equation i=exp(i(pi/2))=cos(pi/2)+isin(pi/2).

  • @onlineMathsTV

    @onlineMathsTV

    11 ай бұрын

    Thanks for the comment, we learnt something new from your comment sir. Respect master.

  • @jinglongchu3829

    @jinglongchu3829

    10 ай бұрын

    i= cos(2n pi + pi/2) + isin(2n pi + pi/2) So, Sqrt(i)= cos (n pi + pi /4) + isin(n pi + pi/4) n is odd or even, 2 answers obtained.

  • @paulorenatovargassilva7298

    @paulorenatovargassilva7298

    8 ай бұрын

    Não entendi nada, não entendo inglês, muito menos esta loucura que ele escreveu na lousa , sem pé nem cabeça...

  • @sreejith18
    @sreejith189 ай бұрын

    Fantastic explanation sir 👍

  • @ayandas8959
    @ayandas895911 ай бұрын

    Hi, I solved it in this way: As we have, √i=(i)^(1/2) =(i^2)(1/4) =(-1)^(1/4) =(cosπ+isinπ)^(1/4) By using de moivre's theorem, =(cosπ/4+isinπ/4) So,value of real part of (√i)=1/√2 value of imaginary part of(√i)=1√2. Thank you.

  • @onlineMathsTV

    @onlineMathsTV

    11 ай бұрын

    Nice and wonderful approach, I just learnt something new here. Thanks for dropping this approach sir. You the master!!! Much respect...👍👍👍💖💖💖

  • @Pablo_Str

    @Pablo_Str

    11 ай бұрын

    Also decision must have πk, k=0,1,2...

  • @lateefkareem

    @lateefkareem

    10 ай бұрын

    using the polar coordinate representation gives a faster solution

  • @shahriarmahbub9680
    @shahriarmahbub968010 ай бұрын

    I watched your video before the night of my exam.I wrote down it and for your this video I wrote it correct and the teacher gave me full marks.Thanks

  • @onlineMathsTV

    @onlineMathsTV

    10 ай бұрын

    Wow!!! We are glad the video tutorial actually help you in your exams and we all @onlinemathsTV join in celebrating you on your success/victory in the exams sir. We hope to hear more positive testimonies of this kind from all our subscribers and viewers all over the world for this is our greatest joy and our aim as far this channel is concern. More success in your academics pursuit sir. Thanks for watching and dropping this comment sir. We all @OnlinemathsTV love you dearly ❤️❤️💖💖💕💕💕

  • @ezzatali6737
    @ezzatali67379 ай бұрын

    we have a2-b2=0 2ab=1 then we can get on a2+b2=1 So 2a2=1 a2=1/2 a=1/r2 or -1/r2 b=1/r2 or -1/r2 So r(i)=a+b i r(i)=+or - 1/r2 (1+i)

  • @akinakinlolu2519
    @akinakinlolu251911 ай бұрын

    I believe that a More easier and Logical way to do this is through the use of Euler notation for complex Numbers. I will not prove it here but the fact is that: on a complex plane, i (imaginary axis) is at 90 degree to the real axis and as such a point with value 1 on the imaginary axis can also be written in polar form R(Cos 90 + isin90). I am using 90 degree here, it should be converted to radians (pi/2 ===difficult to write). But because you can go round and some back to the same position, for family of solutions we can add 2piK to the angle. Thus, Square root (i ) = Square root of (R(cos (pi/2+2piK) + iSin(pi/2+2piK)) R = Square root (1) Square root of (Cos(pi/2+2piK) + iSin(90+2piK)) can also be written as e^i((Pi/2+2piK))/2 where k is 0, 1 since we have square root. So you will have two solutions. e^i(pi/4) , e^i(5/4pi)

  • @onlineMathsTV

    @onlineMathsTV

    11 ай бұрын

    Nice procedure @Akin Akinlolu. Love this comment. Thanks and much love sir.❤️❤️💕💕

  • @danielfranca1939
    @danielfranca1939 Жыл бұрын

    Maths is fun with tutor Jakes...

  • @onlineMathsTV

    @onlineMathsTV

    Жыл бұрын

    Thanks @Daniel Franca. We love u...💕💕💕

  • @mathex95
    @mathex958 ай бұрын

    Very nice

  • @ebakichi1773
    @ebakichi177311 ай бұрын

    5:00 I think a^2=b^2⇒a=b is mistaken a^2=b^2⇒a=b or -b

  • @onlineMathsTV

    @onlineMathsTV

    11 ай бұрын

    Thanks for the observation.

  • @arthurjawara6190
    @arthurjawara619010 ай бұрын

    Thanks for the brilliant math. I am a great fan.

  • @onlineMathsTV

    @onlineMathsTV

    10 ай бұрын

    You're very welcome and thanks for the encouraging words of yours. Much love from all of us @onlinemathstv ❤️❤️❤️💖💖💕💕

  • @raminrasouli7565
    @raminrasouli75658 ай бұрын

    Hello, I did it this way. I don't know if I am correct or wrong: sqrt(i) = i^(1/2) i in polar form, = e^(ipi/2 + 2kpi) so; sqrt(i) = (e^(ipi/2 + 2kpi))^(1/2) which is equal to: e^(ipi/4 + kpi) and when I turn this back to cartsian coordinate form, I got the same answer you got at the end. Thank you very much for your very good and helpful videos.

  • @nasrullahhusnan2289
    @nasrullahhusnan228911 ай бұрын

    It simpler to work this way: sqrt(i)=sqrt(2i/2) =[1/sqrt(2)]×sqrt(2i) But sqrt(2i)=sqrt(1+2i-1) =sqrt(1+2i+i²) as i²=-1 =sqrt(1+i)² =1+i Therefore sqrt(i)=(1+i)/sqrt(2)

  • @onlineMathsTV

    @onlineMathsTV

    11 ай бұрын

    Thanks for this approach sir, you the best....👍👍

  • @samueladler9080

    @samueladler9080

    7 ай бұрын

    Are we letting it be equals to a+bi or it is an identity?

  • @2310fossie
    @2310fossie9 ай бұрын

    You make Maths feel like a walk in the park. You have rekindled my enthusiasm in maths after more than twenty years! Thanks to you , I have become obsessed with Maths. You are a godsend.

  • @rameshchandrasingh6351
    @rameshchandrasingh63517 ай бұрын

    Hoooo just woww, I like your teaching stlyle.

  • @techsinc
    @techsinc7 ай бұрын

    i = exp(i*pi/2). sqrt(i) = sqrt(exp(i*pi/2)) = exp(i*pi/2/2) = exp(i*pi/4) = cos(pi/4) + isin(pi/4) = (1+i)/sqrt(2). Just rotate the phasor in the complex plane from 90 degrees to 45 degrees to get the answer.

  • @mallikarjunaraokotikalapud4828
    @mallikarjunaraokotikalapud48284 ай бұрын

    Sooper sir

  • @StephieJoseph-io3wt
    @StephieJoseph-io3wt Жыл бұрын

    Well explained. Thanks Jakes

  • @onlineMathsTV

    @onlineMathsTV

    Жыл бұрын

    Thanks for gaining values from our channel. Much love.....💖💖💖

  • @user-yz3he2jm4o
    @user-yz3he2jm4o11 ай бұрын

    Could have said b^2=a^2 and 1=2ab. So b=a and 2aa=1 or b=-a and 2a (-a) =1. So a=b=positive square root of 1/2 or a=b=negative square root of 1/2.

  • @jessejordache1869
    @jessejordache186911 ай бұрын

    It's the square root of two over 2 plus the square root of two over 2 times i. Isn't it? It's just trig in an argand diagram: chop the angle in half, and normally you'd square root the magnitude, but i behaves like 1. If you know the sin and cosine of 45 degrees, just use i as the coefficient of the sin term.

  • @pknayak93
    @pknayak9311 ай бұрын

    a=-b also possible.

  • @patearuat664
    @patearuat6649 ай бұрын

    I love how he pronounces 'plus'.

  • @HelenBlackG-xn9hf

    @HelenBlackG-xn9hf

    5 күн бұрын

    Hahahaha 😂😢😂😂😂

  • @taquiarazvi785
    @taquiarazvi78510 ай бұрын

    Thanks for sharing.

  • @onlineMathsTV

    @onlineMathsTV

    10 ай бұрын

    You are most welcome and thanks for the encouragement sir. We love you dearly💖💖

  • @opulence3222
    @opulence32225 ай бұрын

    Nice prove bro

  • @user-nm4cf3qm9y
    @user-nm4cf3qm9y7 ай бұрын

    Thanks very much

  • @onlineMathsTV

    @onlineMathsTV

    7 ай бұрын

    You are welcome and thank you for watching our content and appreciating us sir. We love you sir....❤️❤️❤️💖💖💖💖💕💕

  • @OnosuethaJamesJeremiah
    @OnosuethaJamesJeremiah11 ай бұрын

    nice one

  • @onlineMathsTV

    @onlineMathsTV

    11 ай бұрын

    Thanks for stopping by sir. We appreciate you view and comment and we love you sir....💖❤️❤️

  • @user-tx2gp8ml5i
    @user-tx2gp8ml5i8 ай бұрын

    Où bien tu met juste i sous la forme exponentielle on sait que √a = a^(1/2) après tu calcules

  • @sans1331
    @sans1331 Жыл бұрын

    this one’s from memory sqrti=(i+1)/sqrt2 edit: yep, the negative one works too because any negative value squared is the same as its positive counterpart

  • @onlineMathsTV

    @onlineMathsTV

    Жыл бұрын

    Nice simplification @ that. You are good @ what you do...👍👍

  • @sans1331

    @sans1331

    Жыл бұрын

    @@onlineMathsTV rlly? wow, glad to know that since my dream job is actually a mathematician lol

  • @mahmoodansari1956
    @mahmoodansari195611 ай бұрын

    Good

  • @onlineMathsTV

    @onlineMathsTV

    11 ай бұрын

    Thanks for always be there sir. We pledge to put in our best into what we do in order not to disappoint fathers like you who out of their tight schedule still give our work some attention. Know this sir, everyone @onlinemathstv love and cherish you sir....💕💕💖💖💖❤️❤️

  • @pnachtwey
    @pnachtwey11 ай бұрын

    Think about what two unit vectors that can be multipled to get the square root of i. Or rotate i half way yo 1,0. Think of vectors a d rotations. Then the answers are obvious.

  • @onlineMathsTV

    @onlineMathsTV

    11 ай бұрын

    Wonderful. I will do a research work on that sir. Thanks for the suggestion sir. Respect....👍👍👍

  • @abderrahmanbouhima3624
    @abderrahmanbouhima362411 ай бұрын

    a=b ou a=-b

  • @yannozounguet9512
    @yannozounguet95126 ай бұрын

    ❤❤❤

  • @asparkdeity8717
    @asparkdeity8717 Жыл бұрын

    i = e^i(pi/2) sqrt(i) = e^i(pi/4)

  • @onlineMathsTV

    @onlineMathsTV

    Жыл бұрын

    Nice

  • @mikhaildorservice2343

    @mikhaildorservice2343

    11 ай бұрын

    Они еще не знали про экспоненциальную функцию.

  • @indrajitkatira5953

    @indrajitkatira5953

    10 ай бұрын

    √i=e^i(2nπ+π/2)1/2, n=0,1.

  • @mikhaildorservice2343

    @mikhaildorservice2343

    10 ай бұрын

    @@indrajitkatira5953 maths. Ru

  • @basanagoudamallanagoudra5844
    @basanagoudamallanagoudra58447 ай бұрын

    What is i to the power i ?pls

  • @ashokbiswas5294
    @ashokbiswas52943 ай бұрын

    Explain how you came to know that √i= a+ib form😢 And later somewhere calculation may be shorter too rather you show 😢

  • @OscarMorales-wn7ql
    @OscarMorales-wn7ql11 ай бұрын

    Good Morning! Mr. exhibitor of this block: In his presentation there are certain things that do not have their proper foundation and he uses notations that are not used due to their lack of mathematical logic. For example: 1.- (Vi)² = i Do you know that the rooting and potentiation is in the field of complex numbers? rooting here is a multivalent operation, for example: V(-4) ={ 2i, -2i} 2. From the equation Vi =a+bi and obtain a system of equations that you use as a method to answer the question: find Vi You come to affirm that from a² =b² you get a=b that you do not support. 3.- Write that from a² = 1/2 we obtain a= + - V(1/2) this way of writing is outdated and makes no logical sense Do you know what this meant more than two centuries ago when a mathematician introduced it for short and that it should not be used because it is mathematically illogical? There are many people who reproduce incorrect things learned in school and it is necessary to eliminate them. I wish you a good day!

  • @onlineMathsTV

    @onlineMathsTV

    11 ай бұрын

    @Oscar, sir you just opened my eyes to a very vital point here. I will surely unlearn old skills and learn new ones sir. Your point and comment is well received sir. Thanks for dropping this tip sir. We love you sir....💖💖💖

  • @akinakinlolu2519

    @akinakinlolu2519

    11 ай бұрын

    I believe that a More easier and Logical way to do this is through the use of Euler notation for complex Numbers. I will not prove it here but the fact is that: on a complex plane, i (imaginary axis) is at 90 degree to the real axis and as such a point with value 1 on the imaginary axis can also be written in polar form R(Cos 90 + isin90). I am using 90 degree here, it should be converted to radians (pi/2 ===difficult to write). But because you can go round and some back to the same position, for family of solutions we can add 2piK to the angle. Thus, Square root (i ) = Square root of (R(cos (pi/2+2piK) + iSin(pi/2+2piK)) R = Square root (1) Square root of (Cos(pi/2+2piK) + iSin(90+2piK)) can also be written as e^i((Pi/2+2piK))/2 where k is 0, 1 since we have square root. So you will have two solutions. e^i(pi/4) , e^i(5/4pi)

  • @OscarMorales-wn7ql

    @OscarMorales-wn7ql

    11 ай бұрын

    @@akinakinlolu2519 Good Morning! Remember that you are not solving an equation that does not have two solutions, by definition of the radication of complex numbers as we know it is a multivalent operation, so we have in this case that it is a set that has two elements i this is linked to that in the plane complex it is possible to represent the roots of a complex number that points of a circle with center in (0;0) and with a radius of length equal to the module of the complex number to which the root is applied and the index of the root (n) indicates the number of points distributed in such a way that they divide the circumference into arcs of congruent which actually for n >=3 form a polygon inscribed in the circle defined by this circumference. By definition of the radication using the trigonometric form of expressing a complex number (de Moivre) we have that: Vi={ cos( [pi/2 + 2pi k]/2) + i sin( [pi/2 + 2pi k]/2; where k € {0 , 1} } from here we have: Vi={ cos(pi/4)+ i sin (pi/4) ; cos ( 5pi/4)+ i sin5pi/4 } consequently: Vi={ cos(pi/4) +i sin(pi/4); - cos(pi/4) - i sin (pi/4) } . As we can see, they are not solutions of an equation. This way of expressing Vi indicates the multivalence of the root, the square roots of i are two complex numbers. that's all. I wish you a good day!

  • @chaparral82
    @chaparral8210 ай бұрын

    it is trivial using the euler formula

  • @onlineMathsTV

    @onlineMathsTV

    10 ай бұрын

    sure

  • @ioncodreanu5850
    @ioncodreanu585011 ай бұрын

    From a^2 = b^2 ===> |a| = |b|

  • @onlineMathsTV

    @onlineMathsTV

    11 ай бұрын

    👍👍👍

  • @abdoulayesow6627

    @abdoulayesow6627

    6 ай бұрын

    And from 1=2ab we know ab>0 so we can conclude a = b. And finally, a=±√(1/2)

  • @ChristelleHilaire-lb6pu
    @ChristelleHilaire-lb6pu7 ай бұрын

    i=-2

  • @AbiLove-rl7cn
    @AbiLove-rl7cn11 ай бұрын

    ♥️♥️♥️

  • @onlineMathsTV

    @onlineMathsTV

    11 ай бұрын

    Thanks for the love sir/ma and thanks for watching our video clips. We love you 💕💕💕

  • @EmediongUdoekong
    @EmediongUdoekong16 күн бұрын

    please what about the i in eqtn. iv

  • @onlineMathsTV

    @onlineMathsTV

    14 күн бұрын

    There is no iota, i, in eqn. iv. Kindly recheck sir.

  • @aisawaloki1571
    @aisawaloki15718 ай бұрын

    do anyone like me just considered complex plane very firstly?

  • @izzulimanmuhayidin6986
    @izzulimanmuhayidin698610 ай бұрын

    Could someone explains to me why sqrti is equal to a + bi ?

  • @onlineMathsTV

    @onlineMathsTV

    10 ай бұрын

    It is an identity but I will try and make a video on that sir.

  • @user-vd8fe4qd7e
    @user-vd8fe4qd7e4 ай бұрын

    マイナス1

  • @MathewAdeka
    @MathewAdeka2 ай бұрын

    The answer is 1

  • @palavanchayadaiah339
    @palavanchayadaiah3397 ай бұрын

    Doing it wrong

  • @Dbaby52
    @Dbaby523 ай бұрын

    Stop speeding up his lessons. He's a very detailed teacher and I love his method

  • @leetrask6042
    @leetrask604211 ай бұрын

    1+I/sqrt2

  • @onlineMathsTV

    @onlineMathsTV

    11 ай бұрын

    Bravo👍👍👍

  • @ChristelleHilaire-lb6pu
    @ChristelleHilaire-lb6pu7 ай бұрын

    i=1

  • @WamumaJohn

    @WamumaJohn

    Ай бұрын

    I I've u dia.

  • @mathsr23
    @mathsr232 ай бұрын

    Oh,no

  • @andelaoui4172
    @andelaoui41729 ай бұрын

    C’est faux ce que vous faites.deux erreurs monumentales .je vous conseille de réviser les règles de math.

  • @Adeleq2011
    @Adeleq201111 ай бұрын

    😂 ' simple'. This is MADematics not mathematics

  • @onlineMathsTV

    @onlineMathsTV

    11 ай бұрын

    🤣🤣😂😂 you sounded like this because you are a mathematician and you are good @ it sir.

  • @matokurin
    @matokurin3 ай бұрын

    ❤❤❤❤❤❤❤😊😊

  • @abderrahmanbouhima3624
    @abderrahmanbouhima362411 ай бұрын

    a=b ou a=-b