How Do You Solve a Complex Contour Integral with a Non-Standard Path?

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SUMMARY

The discussion focuses on solving the complex contour integral ∫(x³ - iy²)dz along the path defined by z = γ(t) = t + it³ for 0 ≤ t ≤ 1. Participants confirm that it is appropriate to express z in terms of x and y, specifically using z = x + iy. The derivative dz can be computed as dz = γ'(t) dt, where x and y represent the real and imaginary components of γ(t), respectively. This approach provides a clear method for evaluating the integral along the specified path.

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dan280291
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Hi I'm really not sure how to start this question. I could do it if it was in terms of z but I'm not sure if trying to change the variable using z = x + iy is correct. If anyone could suggest a method I'd appreciate it.

∫(x3 - iy2)dz

along the path z= [itex]\gamma(t)[/itex] = t + it3, 0≤t≤1

Thanks
 
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dan280291 said:
Hi I'm really not sure how to start this question. I could do it if it was in terms of z but I'm not sure if trying to change the variable using z = x + iy is correct. If anyone could suggest a method I'd appreciate it.

∫(x3 - iy2)dz

along the path z= [itex]\gamma(t)[/itex] = t + it3, 0≤t≤1

Thanks

I think it's safe to assume that [itex]z = x + iy[/itex] if nothing to the contrary is given.

You have [itex]z = \gamma(t)[/itex] so [itex]dz = \gamma'(t)\,dt[/itex] and [itex]x[/itex] and [itex]y[/itex] are respectively the real and imaginary parts of [itex]\gamma(t)[/itex].
 

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