Integrating exponetial of z over the conjugate of z

[marqeeete]

Im doing some complex variable "counter integration" problems and this one came up.

I = \oint e ^{\frac{z}{\overline{z}}}dz


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:

\frac{z}{\overline{z}} = e^{2i\theta}

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!

 

[Ray Vickson]

Im doing some complex variable "counter integration" problems and this one came up.

I = \oint e ^{\frac{z}{\overline{z}}}dz


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:

\frac{z}{\overline{z}} = e^{2i\theta}

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.

 

[jackmell]

Im doing some complex variable "counter integration" problems and this one came up.

I = \oint e ^{\frac{z}{\overline{z}}}dz 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:

\frac{z}{\overline{z}} = e^{2i\theta}

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:

\mathop\oint\limits_{|u|=1} e^{u^2} du

How about this one then:

\int_0^{2\pi} e^{e^{2it}} ri e^{it} dt

Then what happens if you just substitute z=re^{it} into yours?

 

[marqeeete]

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.
e^{z/\overline{z}}

as

e^{z^2/\left|z\right|^2}

then i get:

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

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

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

\oint_{|z|=1} e^{z^2}dz = 0

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

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

and finally:

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

does it sound reasonable to you??

 

[jackmell]

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.
e^{z/\overline{z}}

as

e^{z^2/\left|z\right|^2}

then i get:

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

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

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

\oint_{|z|=1} e^{z^2}dz = 0

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

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

and finally:

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

does it sound reasonable to you??

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

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

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:

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.