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

[@ekxo1126]

To all of you that are saying this video is incorrect since sqrt is not multiplicative on C: he's talking about skerootoh, not square root, it's completely different

[@AsadInternationalAcademy]

Thank you so much!

[@AsadInternationalAcademy]

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[@bismarckvazdacosta9078]

Sqrt(i)=(i)**1/2 i=exp(ipi/2) Sqrt(i)=exp(i pi/4)=cos(pi/4)+isin(pi/4) and Sqrt(-i)=exp(-i pi/4)=cos(pi/4)-isin(pi/4) Then It gives 2*cos(pi/4)=sqrt(2)

[@AsadInternationalAcademy]

thank you!

[@nasrullahhusnan2289]

i=sqrt(-1) --> i=(-1)^½ -i=-sqrt(-1) --> -i=-(-1)^½ Let x=sqrt(i)+sqrt(-i) Upon squaring x²=i+(-i)+2sqrt[i×(-i)] =2sqrt(-i²) =2sqrt[-(-1)] =2sqrt(1) =2 --> x=±sqrt(2)

[@AsadInternationalAcademy]

nice job, thanks

[@AsadInternationalAcademy]

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[@Leonhard_Euler99]

That's incorrect. In complex world you don't have the property sqrt(ab) = sqrt(a) • sqrt(b), which is used in this video. The function of adding is one-valued. It's just sqrt(2), without -sqrt(2). The additional wrong sol came from using the forbidden property above.

[@mcguides9176]

They are wrong but actually you are also wrong. It is not an extra solution you are getting here, you are instead missing 2 solutions. This expression can equal +-root 2 as well as +-root 2 i. Whenever you take a square root of a complex number it has 2 values, easily derived by using eulers identity. By each of the 2 values we get 4 different values of this expression

[@Leonhard_Euler99]

@@mcguides9176 Sorry, but no. The function f: C --> C given by f(a,b) = a+b is one-valued. You don't talk about computing numbers like here, but about solving equations. Equation z^2 = -2 has two solutions, + - sqrt(2i), but if you want compute sqrt(i) + sqrt(-i), you can't get two different values. It can be computed like this: sqrt(i) = e^(iπ/4) sqrt(-i) = e^(-iπ/4) Hence, sqrt(i) + sqrt(-i) = e^(iπ/4) + e^(-iπ/4) = 2cos(π/4)= sqrt(2). sqrt(i) + sqrt(-i) is real and it's only sqrt(2).

[@AsadInternationalAcademy]

why?

[@AsadInternationalAcademy]

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[@AsadInternationalAcademy]

ok

[@文率]

just drawing i can be shown as a vector which is 1ㄥ90degree and -i is 1ㄥ-90 degree sqrt(1ㄥ90）= sqrt(1)ㄥ90/2=1ㄥ45 degree and sqrt(1ㄥ-90）=sqrt(1)ㄥ-90/2=1ㄥ-45degree plus these are 1ㄥ45 +1ㄥ-45 =sqrt(2)ㄥ0 which is sqrt(2)

[@AsadInternationalAcademy]

okay thanks

[@AsadInternationalAcademy]

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[@key_board_x]

u = i ← this is a complex number The modulus is: 1 → the modulus of √i will be: 1 The argument is: π/2 → the argument of √i will be: π/4 So you can deduce that the roots are: u1 = cos(π/4) + isin(π/4) → then you add an angle of (2π/2), i.e. π u2 = cos{(π/4) + π} + i.sin{(π/4) + π} u2 = - cos(π/4) - i.sin(π/4) v = - i ← this is a complex number The modulus is: 1 → the modulus of √(- i) will be: 1 The argument is: 3π/2 → the argument of √(- i) will be: 3π/4 So you can deduce that the roots are: v1 = cos(3π/4) + isin(3π/4) → then you add an angle of (2π/2), i.e. π v2 = cos{(3π/4) + π} + i.sin{(3π/4) + π} v2 = - cos(3π/4) - i.sin(3π/4) Summary: u1 = cos(π/4) + isin(π/4) = [(√2)/2].(1 + i) u2 = - cos(π/4) - i.sin(π/4) = [(√2)/2].(- 1 - i) v1 = cos(3π/4) + isin(3π/4) = [(√2)/2].(- 1 + i) v2 = - cos(3π/4) - i.sin(3π/4) = [(√2)/2].(1 - i) It gives: u1 = [(√2)/2].(1 + i) u2 = [(√2)/2].(- 1 - i) v1 = [(√2)/2].(- 1 + i) v2 = [(√2)/2].(1 - i) First possibility: √i + √(- i) = u1 + v1 u1 = [(√2)/2].(1 + i) v1 = [(√2)/2].(- 1 + i) ------------------------------------------------------- = [(√2)/2].[(1 + i) + (- 1 + i)] = [(√2)/2].(2i) = i√2 Second possibility: √i + √(- i) = u1 + v2 u1 = [(√2)/2].(1 + i) v2 = [(√2)/2].(1 - i) ------------------------------------------------------- = [(√2)/2].[(1 + i) + (1 - i)] = [(√2)/2].(2) = √2 Third possibility: √i + √(- i) = u2 + v1 u2 = [(√2)/2].(- 1 - i) v1 = [(√2)/2].(- 1 + i) ------------------------------------------------------- = [(√2)/2].[(- 1 - i) + (- 1 + i)] = [(√2)/2].(- 1 - i - 1 + i) = - √2 Fourth possibility: √i + √(- i) = u2 + v2 u2 = [(√2)/2].(- 1 - i) v2 = [(√2)/2].(1 - i) ------------------------------------------------------- = [(√2)/2].[(- 1 - i) + (1 - i)] = [(√2)/2].(- 2i) = - i√2

[@AsadInternationalAcademy]

Wow! great job, thanks

[@AsadInternationalAcademy]

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[@najmaarshad3452]

So much likeable video ❤

[@AsadInternationalAcademy]

thanks

[@lefebvrem4182]

@Leonhard : incorrect indeed, because of a doubtful shortcut, +/-Any +/-Thing is not +/-(Any+Thing), there are actually 4 possible solutions. But this all depends on how one interprets Sqrt(X) in the 2-D complex world. The obvious solution is Sqrt(2), but if you consider Sqrt(X) as "any number which raised to the power of 2 equals X", then there are 4 solutions (not 2) : Sqrt(2), -Sqrt(2), and also i.Sqrt(2) and i.-Sqrt(2). That is obvious with the polar representation.

[@AsadInternationalAcademy]

thank you so much!

[@AsadInternationalAcademy]

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[@Steven-v6l]

by definition sqrt(x) is the principal square root of x -- a single value if x is real, the principal square root is the positive root if x is complex (x = r*e^(iθ)) the principal square root of x is; sqrt(r * (1 + 2/(-1 + coth((i * θ)/2)))) note… coth() is the hyperbolic cotangent

[@thanosZr]

i^(1/2) + (-i)^(1/2) = (e^(i*π/2))^(1/2) + (e^(i*3π/2))^(1/2) = ... = i2^(1/2). This is a correct solution that you missed due to +- calculations after the roots

[@AsadInternationalAcademy]

thank you so much!

[@AsadInternationalAcademy]

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[@SHATOSHI123]

It has 4 ans 2 real 2 imaginary. I remembered doing it on class 11. Its super easy but you made it complex

[@AsadInternationalAcademy]

okay thanks

[@AsadInternationalAcademy]

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[@Steven-v6l]

WHAT IS A SCARE-ROOT ???

[@AsadInternationalAcademy]

square root = 1/2

[@AsadInternationalAcademy]

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[@louisdavidzouna-i3r]

Congratulations !

[@AsadInternationalAcademy]

thank you so much!

[@AsadInternationalAcademy]

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[@padraiggluck2980]

exp(i *π/4) + exp(-i*π/4) = sqrt(2)

[@AsadInternationalAcademy]

Nice job! thanks

[@Steven-v6l]

lemma: for a,b,c real and c>0 √((a+bi)*c) == √(a+bi) * √(c) z = a+bi lz = √ (^2 + b^2) θ = arctan(b/a) z = lz*(cos(θ) + i*sin(θ)) z = lz*e^(iθ) √z = √lz * e^(iθ/2) w = c*(a+bi) w = c*a + c*bi lw = √ ( (ca)^2 + (cb)^2) lw = √ ( c^2 *( a^2 + b^2)) lw = √ ( c^2 )* √( a^2 + b^2)) lw = c * lz θ = arctan(cb/ca) = arctan(b/a) w = lw*(cos(θ) + i*sin(θ)) w = lw * e^(iθ) √w = √lw * e^(iθ/2) √w = √(c*lz) * e^(iθ/2) √w = √(c)*√(lz) * e^(iθ/2) √w = √(c) * √z Q.E.D.

[@AsadInternationalAcademy]

nice

[@AsadInternationalAcademy]

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[@walterwen2975]

Oxford entrance exam question: √i + √(- i) = ? i = (2i)/2 = (1 + 2i - 1)/2 = (1 + 2i + i²)/2 = [(1 + i)/√2]²; - i = (- 2i)/2 = [(1 - i)/√2]² √i + √(- i) = √{[(1 + i)/√2]²} + √{[(1 - i)/√2]²} = ± (1 + i)/√2 ± (1 - i)/√2 = ± 2/√2 = ± √2

[@AsadInternationalAcademy]

Wow! that's great, thanks

[@AsadInternationalAcademy]

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[@thefedor1]

First square it and get i + 2 - i = 2.Then squareroot.

[@AsadInternationalAcademy]

Nice idea, thanks

[@AsadInternationalAcademy]

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[@bismarckvazdacosta9078]

In fact there are more solutions since the exponential is periodic….

[@AsadInternationalAcademy]

okay

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[@omurolmez0]

What a strange number the i is !

[@AsadInternationalAcademy]

i is a part of complex numbers

[@AsadInternationalAcademy]

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[@antontomsic7438]

What a strange teacher thei is. He says, 99 are idiots.