What is the maximum of the function f(z)= e1/z2 in the unit circle?

In summary, the conversation discusses finding the maximum value of f(z) = e^(1/z^2) on the unit circle, which can be simplified to finding an upper limit for |exp[exp(-2 i theta)]| using the hint to split the inner exponential into its imaginary and real parts. By taking the absolute value, it is determined that the upper bound of cos(2t) is 1, which leads to the conclusion that the maximum value of f(z) is e.
  • #1
Firepanda
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s25dts.jpg


So the length of the contour is L(gamma) = 2.pi

and so i have http://images.planetmath.org:8080/cache/objects/7138/js/img1.png

so i need to show max f(z) = e?

So the maximum of f(z)= e1/z2 in the unit circle centre 0, radius 1 implies that 1/z2 should be maximum, and this is when z2 is its lowest possible value.

When this happens f(z) cannot equal e unless the lowest possible value of z is 1, but isn't that the highest possible value?

Any help would be great thanks
 

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  • #2
On the unit circle z = exp(i theta). So you need to find an upper limit for

|exp[exp(-2 i theta)]|

Hint: Split the inner exponential in its imaginary and real parts.
 
  • #3
Count Iblis said:
On the unit circle z = exp(i theta). So you need to find an upper limit for

|exp[exp(-2 i theta)]|

Hint: Split the inner exponential in its imaginary and real parts.

so exp[cos(2t) - i.sin(2t)]

i have an example similar to this stage where i.sin(t) suddenly disappears, so I assume it does that here as well. Could you explain why for me please?

And then the upper bound of cos(2t) = 1 and so the result follows.
 
  • #4
exp[cos(2t) - i.sin(2t)] = exp[cos(2t)] exp[- i.sin(2t)]

And then when you take the absolute value, you use that

|exp(i p)| = 1 for real p.
 
  • #5
Count Iblis said:
exp[cos(2t) - i.sin(2t)] = exp[cos(2t)] exp[- i.sin(2t)]

And then when you take the absolute value, you use that

|exp(i p)| = 1 for real p.

ah i didn't know that, thanks
 

What is a contour integral?

A contour integral is a type of line integral that is performed along a closed curve in the complex plane. It involves integrating a complex-valued function along a path or contour in the complex plane.

Why is contour integration important?

Contour integration is important in many areas of mathematics and physics, including complex analysis, differential equations, and quantum mechanics. It allows us to evaluate complex integrals that cannot be solved using traditional methods.

What are the key properties of contour integrals?

There are several key properties of contour integrals, including additivity, linearity, and the Cauchy-Goursat theorem. Additivity means that the integral over a contour can be split into smaller contours, while linearity means that the integral of a sum is equal to the sum of the integrals. The Cauchy-Goursat theorem states that the integral of a function over a closed contour is equal to 0 if the function is analytic inside the contour.

How do you choose the contour for a specific integral?

Choosing the right contour for a specific integral can be a challenging task. In general, the contour should enclose all the singularities of the integrand and should be chosen such that the integral is easy to evaluate. It is also helpful to choose a contour that takes advantage of any symmetries in the integrand.

What are some common techniques for evaluating contour integrals?

Some common techniques for evaluating contour integrals include the Cauchy integral formula, the residue theorem, and the method of steepest descent. These methods involve using the properties of complex functions and contour integration to simplify the integral and compute its value.

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