Proving Zero Result with Complex Conjugates and Dot Product

Scootertaj
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Homework Statement


Show that the following = 0:
\int_{-\infty}^{+\infty} \! i*(\overline{d/dx(sin(x)du/dx})*u \, \mathrm{d} x + \int_{-\infty}^{+\infty} \! \overline{u}*(d/dx(sin(x)du/dx) \, \mathrm{d} x where \overline{u} = complex conjugate of u and * is the dot product.

2. Work so far
My thoughts: \int_{-\infty}^{+\infty} \! i*(\overline{d/dx(sin(x)du/dx})*u \, \mathrm{d} x + \int_{-\infty}^{+\infty} \! \overline{u}*(d/dx(sin(x)du/dx) \, \mathrm{d} x
=
\int_{-\infty}^{+\infty} \! -i*(d/dx(sin(x)du/dx)*u \, \mathrm{d} x + \int_{-\infty}^{+\infty} \! \overline{u}*(d/dx(sin(x)du/dx) \, \mathrm{d} x

But I don't even know if that's right.
 
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There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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