Evaluating the Integral of a Vector Field Using Cauchy-Schwarz Inequality

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Homework Statement
Prove that ## \left| \int_C f \left(z \right) \, dz \right| \leq \left|f \right|_{max} \cdot L## where ##\left|f \right|_{max}## is the maximum value o ##\left|f(z) \right|## on the contour and ##L## is the arc length of the contour
Relevant Equations
Any complex valued function is of the form ##f \left(z\right) = u \left(x,y\right) + i v\left(x,y\right)##

I think throughout this I need to use the triangle inequality repeatedly

##\left|\vec{a} + \vec{b} \right| \leq \left| \vec{a}\right| + \left|\vec{b} \right|##

Also the Cauchy Schwarz Inequality

##\left| \vec{a} \cdot \vec{b} \right| \leq \left| \vec{a} \right| \left| \vec{b} \right|##

Just for good measure

##\left| f \left(z\right) \right| = \sqrt{u^2 + v^2}##
Here is my attempt (Note:

## \left| \int_{C} f \left( z \right) \, dz \right| \leq \left| \int_C udx -vdy +ivdx +iudy \right|##

##= \left| \int_{C} \left( u+iv, -v +iu \right) \cdot \left(dx, dy \right) \right| ##

Here I am going to surround the above expression with another set of absolute value bars

##\leq \left| \left| \int_{C} \left( u+iv, -v +iu \right) \cdot \left(dx, dy \right) \right| \right| ##

Appealing to Cauchy-Schwarz
##\leq \left| \left| \int_{C} \left| \left( u+iv, -v +iu \right)\right| \left| \left(dx, dy \right)\right| \right| \right| ##

taking the dot product of the newly defined vector field with itself and noting that cross terms vanish

## = \left| \left| \int_{C} \sqrt{u^2 + v^2}\left| \left(dx,dy\right)\right| \right| \right| \leq \left| \left(\sqrt{u^2 + v^2} \right)_{max} \int_{C} \left| \left(dx,dy\right) \right| \right|##

## \leq f_{max} L##
 
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Sorry for the bad LaTeX disaster I think it's fixed now
 
Actually this problem is more trivial than I thought! An absolute value of an integral of a function over an interval (even if the function and interval is complex) cannot be bigger than the max absolute (a real constant) value of the integrand (on the interval/path) integrated over the contour. Since the max absolute value is a constant we can pull it out of the integral. The only thing left in the integrand is the arc length.
 
For the purposes of a rigorous proof, if you write the integral as the limit of a sum, this is just the triangle inequality, plus the absolute value can pass through the limit since it's a continuous function.

Your intuition in your second post sounds right to me.
 
Just some technical points: I assume your f is Analytic, or at least continuous over the contour,. Contour is closed and bounded, therefore contact. That way U,V, and therefore f is guaranteed to have a maximum, as a continuous function defined in a compact set..
 
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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