it's wrong, solution is +/- sqrt2 and +/- i*sqrt2. 4 solutions for combining two +/- options. and also it much easier in polar form "-1=e^(i*pi)" so sqrt(i) =+/- e^(i*pi/4)
@@AndreTewen a = z^2 had 2 solutions, +/- sqrt(a) but z = sqrt(a) only has 1 solution; sqrt(a) because the formula already sais to take the positive only and not the negative. so since the formula is sqrt(i) + sqrt(-i) the only right solution is sqrt(2). It is not possible for this formula to have multiple solutions. See wolframalpha if you dont believe me. how to work out: 1) i = e^i*pi/2 == > sqrt(i) = e^i*pi/4 and sqrt(-i) = e^-i*pi/4 2) since e^i*x = cos(x) + i * sin(x): ==> sqrt(i) + sqrt(-i) = e^i*pi/4 + e^-i*pi/4 = cos(pi/4) + i*sin(pi/4) + cos(-pi/4) + i* sin(-pi/4) = 2*cos(pi/4) = 2* sqrt(2)/2 = sqrt(2)
@@troyvanleuken1304 |C does not have order relationship, and "sqrt" is not à function : we can write sqrt(z) = (sqrt(|z|), arg(z)/2 + k.pi) with k = 0 or 1. So the raised question can be translated into 4 equations following the k values. with the 4 solutions : k=0, k=0. ==> sqrt(2).i k=0, k=1 ==> sqrt(2) k=1, k=0. ==> -sqrt(2) k=1, k=1 ==> -sqrt(2).i
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Lacks the solution: +/-Sqr(2)*i Sqr(i) = (1/Sqr(2))*(1+i) and Sqr(i) = (-1/Sqr(2))*(1+i) And Sqr(-i) = (1/Sqr(2))*(1-i) and Sqr(-i) = (-1/Sqr(2))*(1-i) So, it's necessary to sum all the possibilities, and it supposes four solutions.
If we take as a criterion that the root of a complex number is the complex number with the smallest positive angle in the complex plane, something similar to the fact that the root of a real number is always positive, then there would be a unique solution: Sqr(2)*i But, I'm not sure if this could be a good criterion.
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it's wrong, solution is +/- sqrt2 and +/- i*sqrt2. 4 solutions for combining two +/- options. and also it much easier in polar form "-1=e^(i*pi)" so sqrt(i) =+/- e^(i*pi/4)
Totally agreed ^
Exact.
sqrt(z) if we work with imaginary numbers, owns two values (|z|, arg(z)/2) and
(|z|, arg(z)/2 + pi).
@@AndreTewen a = z^2 had 2 solutions, +/- sqrt(a) but z = sqrt(a) only has 1 solution; sqrt(a) because the formula already sais to take the positive only and not the negative. so since the formula is sqrt(i) + sqrt(-i) the only right solution is sqrt(2). It is not possible for this formula to have multiple solutions. See wolframalpha if you dont believe me.
how to work out:
1) i = e^i*pi/2 == > sqrt(i) = e^i*pi/4 and sqrt(-i) = e^-i*pi/4
2) since e^i*x = cos(x) + i * sin(x):
==> sqrt(i) + sqrt(-i) = e^i*pi/4 + e^-i*pi/4 = cos(pi/4) + i*sin(pi/4) + cos(-pi/4) + i* sin(-pi/4) = 2*cos(pi/4) = 2* sqrt(2)/2 = sqrt(2)
@@troyvanleuken1304
|C does not have order relationship,
and "sqrt" is not à function :
we can write sqrt(z) = (sqrt(|z|), arg(z)/2 + k.pi) with k = 0 or 1.
So the raised question can be translated into 4 equations following
the k values. with the 4 solutions :
k=0, k=0. ==> sqrt(2).i
k=0, k=1 ==> sqrt(2)
k=1, k=0. ==> -sqrt(2)
k=1, k=1 ==> -sqrt(2).i
why/
Cool trick! Bonus points for not losing multiple solutions of sceruto!
Glad you liked it!
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Seems, more convenient to resolve in exponential forn, whe the equation is sgrt (exp (i×pi/2+ 2pi N))+ sgrt (exp(i3pi/2+2pi N)) etc...
thanks
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The answer is +/- sqrt(2). This was surprisingly easy to do!!! I shall use that for practice!!!
okay
I have an easy easy solution: √i+√-i=X --> (√i+√-i)^2=X^2 --> (i+2-i) = X^2 --> X= ±√2. EASY.
wonderful! thanks
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You can't take ±1/√2 as a common factor, since it stands for _either_ plus or minus, and not necessarily at the same time for both terms.
okay but why?
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Lacks the solution: +/-Sqr(2)*i
Sqr(i) = (1/Sqr(2))*(1+i) and Sqr(i) = (-1/Sqr(2))*(1+i)
And
Sqr(-i) = (1/Sqr(2))*(1-i) and Sqr(-i) = (-1/Sqr(2))*(1-i)
So, it's necessary to sum all the possibilities, and it supposes four solutions.
If we take as a criterion that the root of a complex number is the complex number with the smallest positive angle in the complex plane, something similar to the fact that the root of a real number is always positive, then there would be a unique solution: Sqr(2)*i
But, I'm not sure if this could be a good criterion.
I solved to get i sqrt(2).
And multiple AI confirmed my answer. FWIW, I used DeMoivre’s formula to solve.
thanks
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ok
your answer is wrong!!!
why?
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