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
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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)
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)
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.
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
@@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).
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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
@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.
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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
Thank you so much!
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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)
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)
thank you!
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.
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
@@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).
why?
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ok
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
@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.
thank you so much!
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It has 4 ans 2 real 2 imaginary. I remembered doing it on class 11. Its super easy but you made it complex
okay thanks
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exp(i *π/4) + exp(-i*π/4) = sqrt(2)
Nice job! thanks
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
Wow! great job, thanks
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So much likeable video ❤
thanks
Congratulations !
thank you so much!
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First square it and get i + 2 - i = 2.Then squareroot.
Nice idea, thanks
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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
Wow! that's great, thanks
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In fact there are more solutions since the exponential is periodic….
okay
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What a strange number the i is !
i is a part of complex numbers
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