How To Find The Square Root Of Iota, ί. √ί=? | square root of iota, ί.
Vložit
- čas přidán 25. 05. 2023
- 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.
Share this nice video tutorial using the video link below.
• How To Find The Square...
You can equally watch all video tutorials in this channel by clicking the link below.
/ @onlinemathstv
Loving you all for being there all the time. 💖💖💖💖
Also, gain more views and subscribers using promoterskit for free by clicking the link below:
promoterkit.com/tools/youtube...
#proof #iota #squareroot #deductive
Excellent presentation of steps to find square-foot of Iota. Thanks for uploading this video coaching.
5:00
I think
a^2=b^2⇒a=b is mistaken
a^2=b^2⇒a=b or -b
Thanks for the observation.
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).
Thanks for the comment, we learnt something new from your comment sir.
Respect master.
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.
Não entendi nada, não entendo inglês, muito menos esta loucura que ele escreveu na lousa , sem pé nem cabeça...
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.
Nice and wonderful approach, I just learnt something new here.
Thanks for dropping this approach sir. You the master!!!
Much respect...👍👍👍💖💖💖
Also decision must have πk, k=0,1,2...
using the polar coordinate representation gives a faster solution
Fantastic explanation sir 👍
Maths is fun with tutor Jakes...
Thanks @Daniel Franca.
We love u...💕💕💕
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
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 ❤️❤️💖💖💕💕💕
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)
Nice procedure @Akin Akinlolu.
Love this comment.
Thanks and much love sir.❤️❤️💕💕
Thanks for the brilliant math. I am a great fan.
You're very welcome and thanks for the encouraging words of yours.
Much love from all of us @onlinemathstv ❤️❤️❤️💖💖💕💕
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)
Thanks for this approach sir, you the best....👍👍
Are we letting it be equals to a+bi or it is an identity?
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)
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.
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.
Great work
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.
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.
Well explained. Thanks Jakes
Thanks for gaining values from our channel.
Much love.....💖💖💖
I love how he pronounces 'plus'.
Hahahaha 😂😢😂😂😂
a=-b also possible.
Very nice
Шикарное❤
Hoooo just woww, I like your teaching stlyle.
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.
Où bien tu met juste i sous la forme exponentielle on sait que √a = a^(1/2) après tu calcules
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
Nice simplification @ that.
You are good @ what you do...👍👍
@@onlineMathsTV rlly? wow, glad to know that since my dream job is actually a mathematician lol
Sooper sir
Thanks for sharing.
You are most welcome and thanks for the encouragement sir.
We love you dearly💖💖
Thanks very much
You are welcome and thank you for watching our content and appreciating us sir.
We love you sir....❤️❤️❤️💖💖💖💖💕💕
Nice prove bro
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.
Wonderful. I will do a research work on that sir.
Thanks for the suggestion sir.
Respect....👍👍👍
nice one
Thanks for stopping by sir. We appreciate you view and comment and we love you sir....💖❤️❤️
a=b ou a=-b
What is i to the power i ?pls
Could someone explains to me why sqrti is equal to a + bi ?
It is an identity but I will try and make a video on that sir.
please what about the i in eqtn. iv
There is no iota, i, in eqn. iv.
Kindly recheck sir.
i = e^i(pi/2)
sqrt(i) = e^i(pi/4)
Nice
Они еще не знали про экспоненциальную функцию.
√i=e^i(2nπ+π/2)1/2, n=0,1.
@@indrajitkatira5953 maths. Ru
From a^2 = b^2 ===> |a| = |b|
👍👍👍
And from 1=2ab we know ab>0 so we can conclude a = b. And finally,
a=±√(1/2)
❤❤❤
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!
@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....💖💖💖
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)
@@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!
it is trivial using the euler formula
sure
Good
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....💕💕💖💖💖❤️❤️
do anyone like me just considered complex plane very firstly?
Explain how you came to know that √i= a+ib form😢
And later somewhere calculation may be shorter too rather you show 😢
Stop speeding up his lessons. He's a very detailed teacher and I love his method
♥️♥️♥️
Thanks for the love sir/ma and thanks for watching our video clips. We love you 💕💕💕
マイナス1
i=-2
The answer is 1
Doing it wrong
1+I/sqrt2
Bravo👍👍👍
i=1
I I've u dia.
Oh,no
C’est faux ce que vous faites.deux erreurs monumentales .je vous conseille de réviser les règles de math.
😂 ' simple'. This is MADematics not mathematics
🤣🤣😂😂 you sounded like this because you are a mathematician and you are good @ it sir.
❤❤❤❤❤❤❤😊😊
a=b ou a=-b