Integrating exponetial of z over the conjugate of z

In summary: You might be able to solve it with a residue theorem or cauchy's theorem, but those are not really easy to come by.
  • #1
marqeeete
2
0
Im doing some complex variable "counter integration" problems and this one came up.

[itex]I = \oint e ^{\frac{z}{\overline{z}}}dz[/itex]


the integral must be done over a circle with radio r


My first attempt was to do it in the exponetial form, so we have this:

[itex]\frac{z}{\overline{z}} = e^{2i\theta} [/itex]

but when i do so, i get an e to the power e to the power something and i start getting some long equation so i figure out, there must be an easier way to solve the problem. Hope someone can give me a hand. =D
PS: sorry for my english, it´s my second language!
 
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  • #2
marqeeete said:
Im doing some complex variable "counter integration" problems and this one came up.

[itex]I = \oint e ^{\frac{z}{\overline{z}}}dz[/itex]


the integral must be done over a circle with radio r


My first attempt was to do it in the exponetial form, so we have this:

[itex]\frac{z}{\overline{z}} = e^{2i\theta} [/itex]

but when i do so, i get an e to the power e to the power something and i start getting some long equation so i figure out, there must be an easier way to solve the problem. Hope someone can give me a hand. =D
PS: sorry for my english, it´s my second language!

All I can suggest is that you struggle with trying to do the integral! The final answer is very simple, but I cheated and used a computer algebra package to do it.
 
  • #3
marqeeete said:
Im doing some complex variable "counter integration" problems and this one came up.

[itex]I = \oint e ^{\frac{z}{\overline{z}}}dz[/itex] the integral must be done over a circle with radio r My first attempt was to do it in the exponetial form, so we have this:

[itex]\frac{z}{\overline{z}} = e^{2i\theta} [/itex]

but when i do so, i get an e to the power e to the power something and i start getting some long equation so i figure out, there must be an easier way to solve the problem. Hope someone can give me a hand. =D
PS: sorry for my english, it´s my second language!

Can you solve it with the Residue Theorem or Cauchy's Theorem? Well, how about this one:

[tex]\mathop\oint\limits_{|u|=1} e^{u^2} du[/tex]

How about this one then:

[tex]\int_0^{2\pi} e^{e^{2it}} ri e^{it} dt[/tex]

Then what happens if you just substitute [itex]z=re^{it}[/itex] into yours?
 
Last edited:
  • #4
Thanks for the answers, they really got me thinking and I just came up with an idea; now i do not know if its valid or logical in all its statements, but maybe you can tell me if it is..

I can rewrite.
[itex]e^{z/\overline{z}}[/itex]

as

[itex]e^{z^2/\left|z\right|^2}[/itex]

then i get:

[itex]\oint_{|z|=r} e^{z^2/\left|z\right|^2}dz[/itex]

which is equal to
[itex]\oint_{|z|=r} e^{z^2/r^2}dz[/itex]

and when r=1 (in the unit circle) the integral is equal to zero(as it is an holomorfic/analytic function)

[itex]\oint_{|z|=1} e^{z^2}dz = 0[/itex]

then, as it is also holomorfic/analytic between |z|=1 and |z|=r the result must be the same because of the deformation theorm (its a rubber sheet geometry) so:

[itex]\oint_{|z|=r} e^{z^2/\left|z\right|^2}dz=0[/itex]

and finally:

[itex]\oint_{|z|=r} e^{z/\overline{z}}dz=0[/itex]

does it sound reasonable to you??
 
  • #5
marqeeete said:
Thanks for the answers, they really got me thinking and I just came up with an idea; now i do not know if its valid or logical in all its statements, but maybe you can tell me if it is..

I can rewrite.
[itex]e^{z/\overline{z}}[/itex]

as

[itex]e^{z^2/\left|z\right|^2}[/itex]

then i get:

[itex]\oint_{|z|=r} e^{z^2/\left|z\right|^2}dz[/itex]

which is equal to
[itex]\oint_{|z|=r} e^{z^2/r^2}dz[/itex]

and when r=1 (in the unit circle) the integral is equal to zero(as it is an holomorfic/analytic function)

[itex]\oint_{|z|=1} e^{z^2}dz = 0[/itex]

then, as it is also holomorfic/analytic between |z|=1 and |z|=r the result must be the same because of the deformation theorm (its a rubber sheet geometry) so:

[itex]\oint_{|z|=r} e^{z^2/\left|z\right|^2}dz=0[/itex]

and finally:

[itex]\oint_{|z|=r} e^{z/\overline{z}}dz=0[/itex]

does it sound reasonable to you??

Outstanding. Better than my way. But you could have stopped at just:

[tex]\oint_{|z|=r} e^{z^2/r^2}dz=0, \quad r>0[/tex]

since it's holomorphic. And also, not hard to numerically verify it in Mathematica in case you need to check something else that's not so obvious:

Code:
In[1]:=
NIntegrate[Exp[z/Conjugate[z]]*I*Exp[I*t] /. 
   z -> Exp[I*t], {t, 0, 2*Pi}]

During evaluation of In[1]:= NIntegrate::ncvb:NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in t near {t} = {1.1269*10^-7}. NIntegrate obtained -4.996*10^-16-1.27676*10^-15 I and 3.849598995114243`*^-12 for the integral and error estimates. >>

Out[1]=
-4.996003610813204*^-16 - 1.27675647831893*^-15*I

That's basically zero.
 

FAQ: Integrating exponetial of z over the conjugate of z

What is the purpose of integrating exponential of z over the conjugate of z?

The purpose of this integration is to evaluate complex functions and solve problems in areas such as physics, engineering, and economics. It allows us to analyze the behavior of exponential functions in the complex plane and understand their relationship with the conjugate of z.

How do you solve an integral involving exponential of z over the conjugate of z?

The first step is to rewrite the integral in terms of real and imaginary components, using z = x + iy. Then, we can use techniques such as substitution, integration by parts, or contour integration to solve the integral. The final result will be a complex number, which can be converted to its real and imaginary parts if needed.

What is the significance of the conjugate of z in this integration?

The conjugate of z plays a crucial role in this integration as it allows us to manipulate and simplify complex expressions. It also helps us to express complex functions in terms of real and imaginary parts, making them easier to integrate.

Are there any specific conditions for integrating exponential of z over the conjugate of z?

Yes, the function being integrated must be analytic in the region of integration. This means that it must be differentiable and have a unique complex derivative at every point in the region. If this condition is not met, the integral may not exist or may give incorrect results.

What are some applications of integrating exponential of z over the conjugate of z?

This type of integration has many practical applications, including solving differential equations involving complex functions, calculating areas and volumes in the complex plane, and evaluating complex integrals in physics and engineering. It also has applications in signal processing, control theory, and probability theory.

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