Can you solve this? | iota maths problem | Oxford entrance exam question

[@actuariallurker9650]

Use polar coordinates and solve the problem in 3 lines - what a waste of time

[@rohei1681]

Base knowledge of complex analysis

[@TmyLV]

@@rohei1681 YES!! The guy presenting is REALLY STUPID!!

[@pdfads]

And the way they wrote the final answer makes it look like there are four solutions. Should have written +-(1+i)/sqrt(2) , to avoid confusion.

[@CaspaB]

Agree. I did it in my head, thanks to polar coordinates, learnt 50+ years ago.

[@offgrid-bound]

Yup…

[@MrAlvaroxz]

It's eassier if you use the polar form: i=e^[(π/2)+2kπ]i i^½=e^(π/4+πk)i i^½=cos(π/4+πk)+i×sin(π/4+πk) Now if k is even, then: i^½=sqrt(½)+sqrt(½)i If k is odd, then: i^½=-sqrt(½)-sqrt(½)i Then: i^½=±sqrt(½)±sqrt(½)i

[@MathBeast.channel-l9i]

Nice Approach 👍

[@dansf2]

that's pretty nice

[@KipIngram]

Yes, this is generally true, but in this case the conversion to rectangular is trivially easy "in-head" work.

[@bhobba]

That's how I did it.. Euler strikes again. It's like i^i

[@thomasgreene5750]

Right approach, but the two solutions are +sqrt(1/2)*(1+i) and -sqrt(1/2)*(1+i)

[@medboker]

Your solution gives the impression there are 4 solutions where as there only 2 solutions. You have to notice that an and b have the same sign. And therefore leaving only two solutions

[@rvqx]

So the answer is: √i = ±( √2/2 + i√2/2 ) a and b have the same sign since 4:25 2ab=1

[@rainerzufall42]

@@rvqx Indeed, the other two square to -i because of 2ab=-1. But only one of them is the principal root, that is √i = 1/2 √2 + i/2 √2

[@rainerzufall42]

In other words: √i = cos(45°) + i sin(45°)

[@rainerzufall42]

Or even √i = (1 + i) / |1 + i|

[@ManjulaMathew-wb3zn]

Since 2ab is positive an and be should have the same sign. Also since an and b are real the complex 4th roots are ignored. Having said all that polar form is the way to go.

[@SanePerson1]

i = e^{iπ/2) ⇒ √i = ±exp(iπ/4) = ±(cos(π/4) + isin(π/4)) = ±(1/√2)(1 + i) Introducing the polar form of complex numbers makes this exercise trivial.

[@rainerzufall42]

Only one of these two (the positive one) is a principal root of i!

[@christianlopez1148]

i=exp(i*pi/2), then (i)**(1/2)=exp(i*pi/4)=cos(pi/4)+isen(pi/4)=root(2)/2*(1+i)

[@winstonridgewayhardy]

This is only the solution from the first quadrant. The other one is in quadrant 3: exp(i*5pi/4) or equivalently -root(2)/2(1+i)

[@rainerzufall42]

@@winstonridgewayhardy The principal root of i IS the one in the first quadrant! The value in the third quadrant solves x² = + i, but only x = √i = + (1 + i) / √2 is valid. Just like √2 = + 1.414... only, although (- 1.414...)² = 2 as well! There's a stark difference between y = √x and y² = x. The first form has one solution for y, the second form has 2!

[@e.nicolasleon-ruiz5491]

Show me some real life use for this problem, and I won't call this a pure self-satisfaction exercise.

[@dimetree8496]

Frankly, why it has to go through these many steps? You had a^2-b^2=0 and 2ab=1. From first equation you get a=+-b and from second, you get a^2=1/2, i.e., a=+-root over half.

[@keeteo82]

By the definition of the the sqrt sign : sqrt(z) is THE principal square root of z, which is unique. For example, sqrt(1)=1 not -1, and sqrt(i)=1/sqrt(2) + i*sqrt(2), Though the equation z^2 = i has two solutions.

[@rainerzufall42]

Exactly! Only the solution in the first quadrant IS √i.

[@PlasmaFuzer]

I was thinking the same thing. Not to mention the polar coordinate representation. Using a + bi is so much more cubersome.

[@paulcohen6727]

I saw this title last night and found the answer in a minute or two while I was falling asleep, using de Moirvé's Theorem: i^1/2 = r^1/2 (cis @)^ 1/2 or 1^1/2(cos 1/2(cos 90/2 + i sin 90/2) = sqrt(2)/2 + i sqrt(2)/2. All this spinning 0f your algebraic wheels is a waste of time when you can get the answer directly, visualizing the unit circle.

[@osmanhussein3893]

Exactly! That is what i did without even using a pen and paper!

[@briansauk6837]

Indeed. Just imagine how long i^i would take using some convoluted approach like this...

[@mieses-te9yl]

A little bit emphatic, no?

[@lesbbsayan7212]

if a²-b²=0 we have (a+b)(a-b)=0 so a+b=0 or a-b=0 we get a=-b or a=b so 2ab=1 is 2a²=1 or -2a²=1 a²=1/2 or a²=-1/2 but if a is not a complex number we can't have a²=-1/2 so a+b is not 0 and this is a-b=0 a=sqrt(1/2)=+/-1/sqrt(2) so b=+/-1/sqrt(2) we have sqrt(i)=+/-(1/sqrt(2)+i/sqrt(2)) i think this is better than use 2ab=1 because we don't have to use fraction sum.

[@bookert2373]

This video could be used in high school as a teaching tool to illustrate the hard way to solve this problem and then show the way a mathematician would easily solve it using the fact that angles add when multiplying complex numbers. I think Oxford would be looking for the second approach.

[@okaro6595]

What angle when doubled produces 90 degrees? Squaring doubles the angle.

[@jmguiche9234]

It is really easy for an Oxford exam !

[@KipIngram]

Of course - that can be done in one's head in seconds. It's sqrt(2)/2 + i*sqrt(2)/2 and -sqrt(2)/2 - i*sqrt(2)/2.

[@rainerzufall42]

Only the first one is the principal root √i. As an Oxford student, we should know that!

[@pnachtwey]

Think of rotating vectors. I is a vector of length on pointing up at 90 degrees so rotate it right 45 degrees. This provides the obvious solution.

[@pbierre]

It's easy if you know that, plotted in the 2D plane, a complex number's sqrt is rotationally 1/2-way back to the positive real axis. Since i points straight up on the positive imaginary axis, rotationally 1/2 way back has to be the composite √.5 + √.5i. If you multiply (√.5 + √.5i)(√.5 + √.5i) using FOIL, the result you get is i. This is the positive square root.

[@BollyFan2-ue6cq]

Technically the way you have written the answer it is 4 answers, but really there should only be 2 answers. I think someone in the comments below mentions answers to be +/- (1/sqrt(2)) (1 + i) That is just 2 answers. The combos where a and b have different signs are not valid I assume.

[@RexxSchneider]

The combos where a and b have different signs, i.e. ±(1-i)/√2, are the square roots of -i. However, the radical sign conventionally imples the principal root which would be √(-i) = (1-i)/√2.

[@Matteo-nt3qr]

Sqrt i = cos(pi/2)+1[sin(pi/2)]

[@Jeph629]

If you're a "math" beast, then you know it's a "math" problem. Understanding the finer points of language will generally get you farther than knowing the most esoteric math.

[@winstonridgewayhardy]

I do appreciate your clear and easy to follow format... but yes, a small error that you left both + and - signs within the final answer... since there are only 2 answers which as people below have noted is clear from polar form on the unit circle in terms of rotations from e^(iPi/4) or e^(i5Pi/4) and so: In rectangular form just 2 roots of sqrt(i): sqrt(2)/2 + i sqrt(2)/2 and -sqrt(2)/2 - i sqrt(2)/2

[@kpi6438]

A strange method of finding a solution! By the time such a problem appears (in mathematics, physics...), students already know enough about complex numbers and their exponential representation to solve the problem easily and quickly.

[@laogui2425]

using r*cis(θ) works well, I suggest, and gives at once sin^2(θ)=cos^2(θ) and 2r*sinθcosθ=1 so r = 1 and θ=m*pi/4 with m in (1,3,5,7)

[@RexxSchneider]

Not well enough, it seems. The expression exp(mπi/4) where m ∈ { 1, 3, 5, 7 } represents four values. Two of these (m=1, 5) represent the two square roots of i. The other two (m=3, 7) represnt the two square roots of -i. Only the one where m=1 represents the principal value of the square root of i denoted by √(i). And exp(πi/4) evaluates to (1+i)/√2.

[@topquark22]

Easy if you visualize. No calculation required. Using Euler's formula, i = e^i(i pi/ 4) which is the point on the unit circle at 90 degrees, so sqrt(i) = e^(i pi/8), which is the point on the unit circle at 45 degrees, or (sqrt(2), sqrt(2)). This represents sqrt(2) + i sqrt(2). This is the standard convention, that sqrt() has 1 solution, just as with real numbers. But you could also say another solution is -sqrt(2) - i sqrt(2), which is the point on the unit circle at 225 degrees.

[@harisatya5282]

i=(1/2)(2i) =(1/2)(1² +2×1×i+i²) =(1/2)(1+i)² =[{1/sqrt(2)}(1+i)]² => sqrt(i)=±[{1/sqrt(2)}(1+i)]

[@iran-sweden]

you missed i=(-1/2)(-2i), so we have 4 answeres for this question, two other answers are ±[{1/sqrt(2)}*i(1+i)] =±[{1/sqrt(2)}(1-i)]

[@harisatya5282]

@@iran-sweden 👍 💗

[@gaminghd3950]

Nice video. The calculation could have been shorter as from a^2-b^2=0 follows a=b. Then a=+/- 1/sqrt(2).

[@renesperb]

If you write i = exp[ i π/2] ,then it follows that √ i = exp[ i π/4]= cos π/4 +i sin π/4 = √2/2*(1+i) . Note : there is only one value for √ i .There would be two values if you solve z^2 = i .

[@fernandoderoque1920]

sqrt(i) = i^(1/2) = e^i(π/2+2 nπ)^(1/2) =e^i(π/4+nπ) n in Q

[@IRVINMILLER]

A good understanding of math is illustrated by finding the simplest way of solving a problem. Though some times you want to solve it another way to build your confidence.. Most of your commentators realized i=e^((pi/2)i) and solved it in one or two steps. Try solving i^i with your technique.

[@MathBeast.channel-l9i]

Okay SIR 😊 I will try.

[@gilbertogarbi4479]

Much simpler using Euler's equation.

[@MathBeast.channel-l9i]

How...?

[@eng954]

@@MathBeast.channel-l9i (cospi/2 +isinpi/2)^^1/2 then it can be solved easely by dividng the pi to 2

[@MathBeast.channel-l9i]

@@eng954🆗 👍

[@tommyliu7020]

@@MathBeast.channel-l9ijust realize that square root of -1 is half of the rotation from 1 to -1, so the square root of i is half of the rotation from 1 to i, which correspondents to when theta = pi/4 on the unit circle. There you’ve gotten your answer.

[@Heisemberg08]

Power rule of complex number in the trigonometric form. Z^n = p^n[cos(n*x)+isen(n*x)]. The argument (x) and the module p is easier to identify, once that z = i , p=1 and x= pi/2 (90 degrees), with n=1/2. Just substitute on the trigonometric form and it's done. Dont need to use all that way in the algebraic form. The answer is √2/2 +i√2/2. The answer for (-i)^1/2 is √2/2 - i√2/2. Don't forget the odd/even function rule for cos and sin.

[@seoul-13]

Thank you interesting your solution

[@MathBeast.channel-l9i]

@@seoul-13 You are Welcome 🤗

[@jeanrosw]

i represents a vector in the plan xy having coordinates (0,1) racine of i is a vector having coordinate s racine(2)/2, racine(2)/2

[@AkiraNakamoto]

I use the geometric approach to solve the problem in 30 seconds. i is actually THE OPERATION of counter-clockwise rotation of 90 degrees on the complex plane. sqrt(i) is actually THE OPERATION such that you do sqrt(i) twice, it will rotate counter-clockwisely 90 degrees. This corresponds to two solutions: the 45-degree CC rotation (i.e., 1/sqrt(2) + i/sqrt(2)) and the 215-degree CC rotation (i.e., -1/sqrt(2) - i/sqrt(2)). There are only these two possible solutions on the complex plane. Therefore, your answer giving 4 solutions is not right.

[@AkiraNakamoto]

typo: 215 -> 225.

[@MathBeast.channel-l9i]

Boss🫡 You logic is😇

[@steve_s9412]

Yes, that's how I did it. Once you realise that the required operation is a 45 degree rotation then it's obvious from pythagoras that the real and imaginary parts are both root 2 over 2.

[@Heisemberg08]

Negative part in power of complex numbers just appear if π/2

[@marksteers3424]

The e to the i theta is simplest but even in your workings - when you get a2-b2 = 0 and we know that a and b are real then a = b. 2ab =1 becomes 2a2 =1 a = 1/ sqroot 2.

[@ExpressStaveNotation]

root of i = / (unit vector at 45 degrees). = (1 + i)/r2

[@shaileshdhuri4166]

e^(pi *- i / 4 ) = (-1)^(1/4) == i ^ 0.5. One step solution

[@martinrosenau478]

Again the same mistake I already commented in another video this week: √4 is not ±2 (although (-2)² is also 4) but √4 is defined as +2! This is true for all complex numbers (including real numbers): With exception of the value x=0, there are exactly two values y that satisfy y²=x, but only one of the two values is defined as the square root √x. According to the German language Wikipedia, it is the value with the positive real part (for this reason, √4 is +2 (which is equal to +2+0i) and not -2 (which is equal to -2+0i)); and if the real part of the two values y is zero, it is the value with the non-negative imaginary part (for this reason, √(-1) is +i but not -i). For this reason, √i is only (√2/2)(+1+i) but not (√2/2)(-1-i).

[@KipIngram]

They're both square roots. I've always been taught to call the one you identified the "principle square root."

[@martinrosenau478]

@@KipIngram Because language may play a role here, I also checked the English language Wikipedia: If I understand the Wikipedia text correctly, both "+√x" and "-√x" are called "square roots" in English language; however, only one of the two values can be written as "√x" while the other one must explicitly be written as "-√x". I don't know about the original Oxford question, but in the video the question is: "Find √i", not "Find the square root of i".

[@radupopescu9977]

@@KipIngram Indeed. In real life we choose the positive one, because of practical purposes. In Pythagoras theorem you can have negative lengths. But with complex number, we must take all into account. So x^(1/n), has exactly n values, where n is any natural number. (-16)^(1/2) has 2 values: 4i (principal) and -4i (-27)^(1/3) has 3 roots: -3, (3/2 + (3*((3)^(1/2))*i)/2) and (3/2 - (3*((3)^(1/2))*i)/2)

[@tontonbeber4555]

​​@@radupopescu9977x^(1/n) is an expression, a mathematical operation. It as only one value ... By the way, following your "theory", how many values has x^pi ?

[@radupopescu9977]

​@@tontonbeber4555 I did a mistake when I wrote this comment and now I rewrite it, and I apologize for that. Let's take 4^pi= 4^(-i*ln(-1)). And complex log is multivalued! Of course the principal value is 4^pi=77.88.... of 4^(-i*ln(-1)) in practice we use this value, but the 4^(-i*ln(-1)) has INFINITE VALUES. Because Ln(-1) is +/-i*pi; +/-i*3pi, and so on... Do not restrict to reals! Some time thinking only to complex numbers is also a narrow vision. Bicomplex numbers offers new classes of solutions. I like bicomplex numbers because they are commutative in contrast with quaternions which are not. They also have a disadvantage: zero divisors! Let's give you another e.g. where complex numbers are a narrow way to see things. For solving this equation we need bicomplex numbers. Imagine how multi-complex numbers even more diverse. An e.g. of equation which has other solution then 0. x*(i-j)=0, and x is NOT 0. What is the value of x? In bicomplex numbers we define: i^2=-1, j^2=1, i*j=j*i, and i is different of j. So bicomplex numbers are of form: a+bj, where a and b are complex numbers. In this case the solution is x=i-j In bicomplex numbers any number of form a*(1+ij)(1-ij)=0 (zero divisors), where a is any complex number, except 0. So it is possible to have x*y=0, with x and y not 0!

[@phongsakbuala2483]

i^0.5 = e^(i(pi/4 +/- npi))= +/-(1+i)/sqrt2

[@maamouhinda7722]

You way it's ok but too hard and long. Short solution : i= e^(pi/2 +2 k.pi) i^1/2= e^(pi/4 + k.pi)=√2/2 +i.√2/2 for k=2.p and -√/2 -√2/2 for k= 2.p+1

[@johnvonhorn2942]

I'm going to "shoot in the dark" and say that. by definition i^2 = -1 so i = sqr(-1) so sqr(i) = -1 ^ (1/4) That's as good as I get and if it's not good enough then, "who cares?" cos I'm not Leonard Euler or Frederick "Wilhelm" Guass. I work at Poundland on the minimal wage and I'm even pretty terrible at that job.

[@iran-sweden]

i=e^((2πn+π/2)i)=>i^(1/2)=e^((πn+π/4)i)=cos(πn+π/4)+isin(πn+π/4)=±(√2/2)(1 ± i) , n=1,2,3,4,...

[@yairidan5007]

= + - (cos 45 + i sin 45)

[@Electronics4Guitar]

I did this one in my head in about 30 seconds.

[@MathBeast.channel-l9i]

How...? Explain a bit...

[@CaspaB]

Me too. Polar coordinates.

[@RexxSchneider]

@@MathBeast.channel-l9i Recall that i = exp(πi/2). So √(i) = exp(πi/4) = cos(πi/4) + i.sin(πi/4) = 1/√2 + i/√2 = (1+i)/√2. I think 30 seconds is more time than is needed for that.

[@15_sayandeepkundu_ee28]

I did it in DSP(Digital Signal Processing) in Engineering

[@wmatos]

Thank you.

[@MathBeast.channel-l9i]

@@wmatos You are Welcome 🤗

[@srinivasch-re2oq]

√i = 1 I ^4/4 which means i^2 = -1 and then square equal to 1 Then √1 = 1 So √i = 1

[@davesthinktank]

Plot in polar coordinates, take half, answer is obvious with almost zero work.

[@moebadderman227]

let a + bi = √i then (a + bi)² = i, so (a² - b²) + 2abi = i therefore ab = ½ so _a_ and _b_ must have the same sign, and a² = b², so a = b = ±√½ so the two solutions are √i = √½ + (√½)i, and -√½ -(√½)i

[@69Hauser]

A eso le llamo yo simplificar... y malgastar rotulador. Al mismo resultado se llega por vías mucho más simples. Por cierto, menuda tortura de música. Y Einstein, ¿qué tiene que ver con Oxford?. ¡Ah, que lo de Oxford también es un cebo! Vale.

[@EduardoSousaSaraiva-p6i]

Using the exponential format: exp(j(pi/4 + k pi)) k element of Z

[@prajaktab7010]

nice example. unic solution.

[@MathBeast.channel-l9i]

Thankyou so much 😊

[@MathBeast.channel-l9i]

Thanks for your lovely feedback ☺️ It means a lot for us.

[@JunedHussain]

I did not get a^2 - b^2 = 0 and abi=1i. What is the logic behind this

[@dimetree8496]

​@@JunedHussain it means real number equals to real number and imaginary number equals to imaginary number. More conceptually whatever co-efficient an imaginary number has it will never be real number. Think like, 5 apples will never be 3 oranges.

[@MathBeast.channel-l9i]

@@dimetree8496 Appreciable 👍

[@janejohnson8353]

This problem took forever. How long do students have to complete the entrance test? As stated below, there are quicker and easier ways to do it.

[@herbertklumpp2969]

a^2 = b^2 conclude a = + - b Therefore if a=b you get a^2 =1/ 2 a= b = +- sqrt( 1/2) a= - b no solution Result:sqrt,( i) = +- sqrt(1/2)*( 1+i)

[@BigEaster]

Never heard of De Moivre formula? Immediate solution.

[@dariuszb.9778]

It's extremely easy if you know what Euler's identity means and how you do powers and roots on a complex coordinate system.

[@SfhjjFghjki]

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[@rogermarshall2310]

laboured!

[@PaulEhmig]

e^(π/4)i

[@PaulEhmig]

And the others adding 2kπ to pi/4

[@ManojkantSamal]

Respected Sir, Good morning

[@MathBeast.channel-l9i]

Thankyou Good morning too🙂

[@JSSTyger]

√(i) = ±√(2)/2(1+i)

[@Georgios-ft5nm]

More interesting is the fact that i^i is real.

[@michaelalitheturkali]

x* (xlessy)*(xless2y)*(xless3y)*................... and use where useful

[@thomasharding1838]

"Let's Suppose" that i = ±√-1 then can we "suppose" that √i = ± fourth root of -1 ‽

[@itsmetrendy8471]

exactly what I was thinking

[@thomasharding1838]

@@itsmetrendy8471 The problem with our thinking is that it doesn't take 10 minutes to express it.

[@itsmetrendy8471]

@@thomasharding1838 true because I just saw the way how the normal expression just needs another root on the right side and simplified it

[@adgf1x]

i^1/2=(-1)^1/4

[@jsc3417]

Easy Sqart(i)=i^0.5

[@TheOldeCrowe]

i = e^(iπ/2) √i = e^(iπ/4) = cos(π/4) + i sin(π/4) = (1 + i)/√2 Straightforward. 97% failed is a ridiculous click-bait statement.

[@MathBeast.channel-l9i]

Everyone may not be so sharp as you in Mathematics.

[@MYeganeh100]

👌

[@nigellbutlerrr2638]

Wrong, it has one solution. Principal Solution. 1/√2+ i/√2. Five seconds. R. angle theta. Use polar. Five seconds 😂

[@konnischeller5185]

Das ist exakt der Lösungsweg, den ich in der 9. klasse Realschule gefunden hatte.

[@Birol731]

My way of solution ▶ x= √i x²= i i= eᶦθ θ= π/2 ⇒ i= eᶦ⁽π/²⁾ ⇒ x²= cos[(π/2 +2πk)/n] + isin[(π/2 +2πk)/n] n= 2 a) for k= 0 x₁= cos[(π/2+0)/2] + isin[(π/2+0)/2] x₁= cos(π/4) + isin(π/4) cos(π/4)= √2/2 sin(π/4)= √2/2 ⇒ x₁= √2/2 + i √2/2 x₁= √2/2(1+i) b) for k= 1 x₂= cos[(π/2+2π)/2] + isin[(π/2+2π)/2] x₂= cos(5π/4) + isin(5π/4) cos(5π/4)= -√2/2 sin(5π/4)= -√2/2 ⇒ x₂= -√2/2 - i √2/2 x₂= -√2/2(1+i) 𝕃= { x ∈ ℂ : { -√2/2(1+i) , √2/2(1+i) }

[@victorfinberg8595]

trivial if you know complex numbers: a) (1+i)/SQRT(2) b) multiply above by -1 i did that in my head, in seconds. impossible if you don't know complex numbers. problem is, is this something that is taught in high school? if yes, then ALL stem students will do it, and none of the others. so your 97% figure is suspect

[@victorfinberg8595]

p.s. why do this in such an unnecessarily complicated way?

[@antoniosofal7229]

Gostei

[@daniel_feiglin]

Trivial! Draw a picture: sqrt(i) is clearly (1+i)/sqrt(2). It's simple geometry.

[@alicengiz1861]

On the middle of the way it was already obvious that a=b and = +-1/2^0.5. you made a lot of redundant operations

[@carystrunk5771]

Excellent! Utterly fascinating! I also like the fact that you do not speak during the explanations. In addition, your handwriting is pretty.

[@MathBeast.channel-l9i]

Thank you very much! For your excellent feedback ❤️

[@MathBeast.channel-l9i]

Thanks for praiseing my handwriting. A positive feedback from audience means a lot for a content creator. It encourages and boosts the ability to provide a more better work. Thankyou Boss😊 You are great God bless you 😇

[@Herby-1620]

The comments already entered indicate that SQRT(i) has at most 2 solutions, not 4. The problem I see is that the result is not simplified properly. There should be no radicals in the denominator. 1/sqrt(2) is properly written as sqrt(2)/2. This was not done!

[@Arcangelnino]

but why is a^2 - b^2=0 and i^1/2= a + bi? these are only supositions and theres no way to prove it, im new to complex numbers so tell me if im wrong plz

[@colinmccarthy7921]

If the square root of i = ?.You could say the square root of 1 = 1.This should help.

[@Mesa_Mike]

The square root symbol only wants the principal square root (i.e. the root with the non-negative Real component). So there is only one answer: √i = √2/2 + i√2/2

[@videofountain]

Please see the answer above yours.

[@davidwright8432]

Taking only the principal squrt is a convention, not logically mandated. It's not wrong to deal with both; just a violation of convention, not logic.

[@ЧинЗан-в3э]

Бесконечность, это функция, которая позволяет добавит ещё 1 тогда, это превращается в сигнатуру, а не бесконечность. Умнийчайие сами во славу перемещения во времени..

[@RealQinnMalloryu4]

(x ➖ 1ix+1i)

[@somersetcace1]

I know this is probably just laymen's ignorance, but this is what I never understood about advanced math. You have a problem to solve for and the answer is even more convoluted than the initial equation. It's doesn't appear you solved for the square root of i, but rather went through a lengthy process of restating the equation in a different way. We still don't know the value of `i,` or its square root. So, what's the point of it, other than a way of testing ones ability to rearrange equations, and/or apply mathematical theorems? I suppose that's fine for a college entrance exam, but I see no practical application.

[@moonshade9860]

This equation probably has no practical application since it’s just something to practice the concept with. The concept itself is used a whole lot in practical applications. Digital telecom communications rely on it. Without it there is no gsm or any mobile network.

[@somersetcace1]

@@moonshade9860 Thanks for the response and that makes sense. I do use mathematical formulas in my own job, but they're preset formulas, and all I'm really doing is plugging in the variables and doing the math. There might be some rearrangement depending on what variable I'm looking for, but that's about it. This, on the other hand, is a completely different level of application and clearly over my head. lol

[@gnanadesikansenthilnathan6750]

Nice

[@AnilSonkar-ly5ex]

Nice video

[@MathBeast.channel-l9i]

@@AnilSonkar-ly5ex it means a lot for us❤️

[@T121T]

Is that four answers or two? Tye plus and minus parts?

[@richardmullins44]

on the argand diagram i is (1, pi/2), so sqrt of i is (1, pi/4). unfortunately my brain fades out at this point, and I can't think what the other value is? Would it be (1, pi/4 + pi)?

[@lynnrathbun]

solve in 5 seconds in polar coordinates, in your head

[@joeviolet4185]

I purport that out of the 3% who solved the problem, 97% again never heard about the complex number plane and how to do algebra on it. Thus, the simple solution is: Divide the angle the complex number 0+1·i forms with the positive real axis, which is 90° by 2 and take the square root of its absolute value, which is 1. Next draw the resulting complex number a+b·i, which then forms an angle of 45° and has an absolute value of 1. If you remember a little bit of what you learnt about sin and cos at school, you immediately know that a=b=1/sqrt(2), otherwise look it up in any formulary.

[@henkn2]

Can you demonstrate what happened in your head in those 5 seconds?

[@IllllllllllIIlIIIIlIlllI]

I thought his answer was (-1)^1/4 😅

[@Imran52Feb]

- 1^1/4 should be acceptable too

[@amehachewa4757]

You did a long trip to those who don't know mathematics, so please don't make confusion. Mathematics is fun,so make it simple by using the formula as we have learned.

[@MathBeast.channel-l9i]

Explain a bit... How...? Which Formula...?

[@rainerzufall42]

The result is plain wrong! Only 2 of these 4 complex numbers are a solution (the other 2 square to -i = 1/i). The only principal root of i is ++ => sqrt(i) = 1/sqrt(2) + i/sqrt(2) = 1/2 sqrt(2) + i/2 sqrt(2)

[@Etothe2iPi]

That is so painful! It took me 5 seconds to see the two solutions of x^2=i. Why do yo write a^2 with just one set of parentheses. Why not 2, or 3?? Surely a^2= (((a)))^2.

[@Christopher-e7o]

+5:)

[@MathBeast.channel-l9i]

🤔

[@RexxSchneider]

No. The convention is that √(i) denotes the principal value of the square root of i, which is (1+i)/√2. i = exp(πi/2) so √(i) = exp(πi/4) = cos(π/4) + i.sin(π/4) = 1/√2 + i/√2 = (1+i)/√2. The other square root of i is just -(1+i)/√2. There can only be two values for any square root, not the four that you imply. And please quit the ridiculous click-bait. It's not an "Oxford entrance exam question", whatever you imagine that to be.

[@PauperRulesCommittee]

i was the moment my brain refused to learn any more maths

[@MathBeast.channel-l9i]

Which moment...?

[@PauperRulesCommittee]

​@@MathBeast.channel-l9iimaginary numbers like i. I just couldn't get over the fact we agreed not to square root a negative number, and then imaginary numbers showed up. I just couldn't

[@debasishbose6376]

Have never seen knowledge so well dramatized.

[@catherinedemorvan5032]

YES

[@josepherhardt164]

Before listening: Did they forget the +2nπi? (AKA, the +c of the complex world.)

[@aljawad]

Use Euler’s identity and Euler’s formula.

[@pennstatefan]

I got -1.

[@philippedelaveau528]

Il me semble qu’un candidat à Oxford ne sait pas résoudre ce problème immédiatement en une ligne par l’exponentielle, c’est qu’il n’à vraiment le niveau et sera illico recalé.