Reverse direction for complex functions

In summary, reverse direction for complex functions involves the process of determining the inverse of a complex function. This can be achieved through various methods, including finding the inverse of each component of the function or using the inverse function theorem. The concept of complex conjugates is also important in this process, as it helps in simplifying the inverse function. Reverse direction for complex functions is necessary for solving equations involving complex numbers, and it is a crucial skill in advanced mathematics and engineering.
  • #1
ognik
643
2
Hi

An exercise asks to show $ \int_{a}^{b}f(z) \,dz = -\int_{b}^{a}f(z) \,dz $
I can remember this for real functions, something like $ G(x) = \int_{a}^{b}f(x) \,dx = G(b) - G(a), \therefore \int_{b}^{a}f(x) \,dx = G(a) - G(b) = -\int_{a}^{b}f(x) \,dx $

I have seen 2 approaches, either parametizing w.r.t t or expanding z=u+iv, what I don't understand is why I need to paramatise/expand f(z)? Both of those approaches also have complex intergrands, so why does the paramatise/expand step make a difference?
 
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  • #2
ognik said:
Hi

An exercise asks to show $ \int_{a}^{b}f(z) \,dz = -\int_{b}^{a}f(z) \,dz $
I can remember this for real functions, something like $ G(x) = \int_{a}^{b}f(x) \,dx = G(b) - G(a), \therefore \int_{b}^{a}f(x) \,dx = G(a) - G(b) = -\int_{a}^{b}f(x) \,dx $

I have seen 2 approaches, either parametizing w.r.t t or expanding z=u+iv, what I don't understand is why I need to paramatise/expand f(z)? Both of those approaches also have complex intergrands, so why does the paramatise/expand step make a difference?

If the integral $\displaystyle \int_{a}^{b} f(z)\ d z$ does not depend on the path connecting a and b in the complex plane, then f(z) is analytic in the whole complex plane and effectively is $\displaystyle \int_{a}^{b} f(z)\ d z = - \int_{b}^{a} f(z)\ d z$...

... otherwise, i.e. if f(z) has some singularities in the complex plane, that may be not true...

Kind regards

$\chi$ $\sigma$
 
  • #3
Thanks χ σ , but the exercise doesn't specify that f(z) is analytic.

Maybe more context would help - it appears just after a section on Cauchys Integral theorem.

That section also says: "A consequence of the Cauchy integral theorem is that for analytic functions the line integral is a function only of its end points, independent of the path of integration," - which would surely mean I could directly use the same argument I used for real functions above, WITHOUT paramatising or expanding, wouldn't it?
 
  • #4
ognik said:
Thanks χ σ , but the exercise doesn't specify that f(z) is analytic.

Maybe more context would help - it appears just after a section on Cauchys Integral theorem.

That section also says: "A consequence of the Cauchy integral theorem is that for analytic functions the line integral is a function only of its end points, independent of the path of integration," - which would surely mean I could directly use the same argument I used for real functions above, WITHOUT paramatising or expanding, wouldn't it?

The difference between integrating in $\mathbb {R}$ and integrate in $\mathbb {C}$ is that in the second case it must be generally specify not only the limits of integration a and b, but also the path that connects a and b...

Let's consider the following illustrative example taking $f(z) = \frac{1}{z}, a=1, b=-1$ and suppose to choose as path of integration the upper half cirle of the figure...
http://cdn.simplesite.com/i/c2/c2/286260055773790914/i286260064307432655._szw1280h1280_.jpg

Setting $\displaystyle z=e^{i\ \theta}$ we have...

$\displaystyle \int_{1}^{-1} \frac{d z}{z} = i\ \int_{0}^{\pi} d \theta = i\ \pi\ (1)$

Now we have to integrate from -1 to 1 and, among other, we have the opportunity to choose two different paths...

a) the upper half circle in the figure, i.e. the reverse path than the previous case, and we have...

$\displaystyle \int_{-1}^{1} \frac{d z}{z} = i\ \int_{\pi}^{0} d \theta = - i\ \pi\ (2)$

... so that Your relation is verified...

b) the bottom half circle of the figure and we have...

$\displaystyle \int_{- 1}^{1} \frac{d z}{z} = i\ \int_{\pi}^{2\ \pi} d \theta = i\ \pi\ (3)$

... and Your relation isn't verified...

The reason of that is that the function $f(z)= \frac{1}{z}$ is not analytic in the whole complex plan, having a singularity in z=0...

Kind regards

$\chi$ $\sigma$
 
  • #5
Thanks χ σ ,
I know that if a function is complex analytic, it is independent of the path - so it seems to me I have no other option but to assume F(z) is analytic?

My concern is that the solutions I have seen either expand the function (u,v) or parametise it (t) - and then conclude that the reverse integral does change sign. But my logic says - if the function is not analytic, then neither expanding nor parametising will suddenly make it analytic?
 

FAQ: Reverse direction for complex functions

What is reverse direction for complex functions?

Reverse direction for complex functions refers to the process of finding the inverse of a complex function. This involves switching the roles of the input and output variables to solve for the original input value given a known output value.

Why is it important to understand reverse direction for complex functions?

Understanding reverse direction for complex functions allows us to solve for the original input value in situations where we only know the output value. This is especially useful in engineering and scientific applications where complex functions are commonly used.

How do you find the inverse of a complex function?

To find the inverse of a complex function, you need to switch the input and output variables and solve for the original input value. This can involve using algebraic techniques such as factoring, completing the square, or using trigonometric identities.

Are there any restrictions for finding the inverse of a complex function?

Yes, there are certain restrictions for finding the inverse of a complex function. The function must be one-to-one, meaning that each input value corresponds to a unique output value. Additionally, the function must be invertible, meaning that it can be reversed without losing any information.

Can all complex functions be reversed?

No, not all complex functions can be reversed. As mentioned earlier, the function must be one-to-one and invertible in order for its inverse to exist. Some complex functions, such as exponential and logarithmic functions, have well-defined inverse functions, while others do not.

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