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Let y be the curve in X defined

y(m)=(m^3/3)-m+(i*m^2), m e [-2,2]

a)Let f(z) = -i*z. Compute ∫y f(z)dz , (integral y)

b)Calculate the length of y

To avoid problems with notation, let me define the curve in the complex plane as

Ct) = t^3/3 - t + it^2.

Defining (as usual) z = x +iy and dz = dx + idy gives f(z) = -iz = -i(x+iy) and

∫f(z)dz = -∫[i(x+iy)(dx+idy) = ∫[(-ix+y)dx+(x+iy)dy]

Associating for the moment P = -ix+y and Q = x+iy we can use a theorem that says that ∂P/∂y = ∂Q/∂x (= 1) means that the integral is path independent. We thus only need the values of the curve at the endpoints

C(2) = 2/3 + 4i and C(-2) = -2/3 +4i so that

∫f(z)dz = ∫izdz = iz^2/2 evaluated at the above limits. I get ∫f(z)dz = -i16/3.

b) The length of C(t) is given by the usual expression for arc lengths. Defining the real part x = t^3/3 - t and imaginary part y = t^2,

S = ∫[(dx)^2+(dy)^2]^1/2 = ∫[(dx/dt)^2+(dy/dt)^2]^1/2dt = ∫[(t^2-1)^2+(2t)^2]^1/2dt = ∫(t^2+1)dt = 2t^3 + t with limits (-2, 2). The answer I get is S = 32.

I filled in most of the steps but you should check my algebra. Let me know if you have any questions. In the future, please show me your work and/or ask a specific question rather than just give me a homework problem.

Randy

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