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Question about complex numbers and laurent series?

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  • Question about complex numbers and laurent series?


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Answer #1 | 23/07 2014 13:07
Not enough explanation.
Positive: 50 %
Answer #2 | 23/07 2014 12:29
i^2 = -1 - i^2 = 1 ok?
Answer #3 | 24/07 2014 02:45
(x^2 - 1) = (x+1)(x-1) right? (z^2 + 1) = (z + sqrt -1)(z-sqrt (-1)) f(x) = 1/(z+1) g(x) =1/(z+i) f(x) = g(x)/(z-i) a_0 = 1/2pi i ∫ g(x)/(z-i) dz if the radius is < 2 this integrates a_0 = g(i) 1/2i = -i/2 a_1 = 1/2pi i ∫ g(x)/(z-i)^2 dz = g'(i) -1/(z+i)^2 = 1/4 a_2 = 1/(2i)^3 = i/4 a_3 = -1/(2i)^4 sum (-i/2)^n (z-i)^n sum ((1-zi)/2)^n
Answer #4 | 24/07 2014 09:45
(x^2 - 1) = (x+1)(x-1) right? (z^2 + 1) = (z + sqrt -1)(z-sqrt (-1)) f(x) = 1/(z+1) g(x) =1/(z+i) f(x) = g(x)/(z-i) a_0 = 1/2pi i ∫ g(x)/(z-i) dz if the radius is < 2 this integrates a_0 = g(i) 1/2i = -i/2 a_1 = 1/2pi i ∫ g(x)/(z-i)^2 dz = g'(i) -1/(z+i)^2 = 1/4 a_2 = 1/(2i)^3 = i/4 a_3 = -1/(2i)^4 sum (-i/2)^n (z-i)^n sum ((1-zi)/2)^n
Positive: 0 %
Answer #5 | 23/07 2014 19:29
i^2 = -1 - i^2 = 1 ok?

Possible answer