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This is a preview of the volume eight of  'Critical Comments ...'  -
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                   C R I T I C A L   C O M M E N T S   O N

                            WHY BOTHER TO COMPUTE?
                              and related papers

                                              
                                P A R T   VIII

                        Complex Variables in Engineering



                              Compiled by: T. Leko
      
--------------------------------------------------

                                  M.  MCCARTY

--------------------------------------------------

Subj: Re: Bracket a complex root
Date: 96-07-08 12:15:18 EDT
From: JMCCARTY@altair.dsccc.com (MIKE MCCARTY -- 93789)
To: Tleko@aol.com
CC: JMCCARTY@altair.dsccc.com

)      Presented at the 1996 GAMM-Conference in Prague
)    
)      Leko, T.
) 
)      COMPLEX VARIABLES IN ENGINEERING TRUE OR FALSE?
) 
)      Abstract: This paper disproves the foremost
)      indispensible fundamental importance of Cauchy's
)      integrals in the theory of analytic functions of
)      a complex variable. In particular, Cauchy's
)      integral theorem, Cauchy's integral formula,
)      the definition of analyticity, Taylor's series
)      and Laurent's series have all escaped the
)      desired scrutiny.
) 

[elementary stuff cut]

)                      4. Three Examples
)
)     Example 1.  The analytic property of:
) 
)                 f(z) = 1/(1-z)^2
)                                           
)                 (R. Mennicken)
) 
)     Example 2.  On a boudary value problem via Poisson's integral:
) 
)                 f(z) = ((1-z)/(pi sqrt z)) ln((1+sqrt z)/(1-sqrt z))
) 
)                 (F. Goerisch)
)
)     Example 3.  On the value of Cauchy's integral:
) 
)                 Int (1/(t-z)) dt
) 
)                 (K. Habetha)  .
)
) 
)                       5. Results
) 
)     The results of the three examples are documented in the seven
)     volumes of 'Critical Comments on Why Bother to Compute? catalogued
)     in the Library of Congress, Washington, DC and several university
)     libraries in the USA and Germany. In their private communications
)     Professors Mennicken, Goerisch and Habetha have verified that the
)     three examples do not represent analytic functions in the unit
)     disk contrary to expectation.
) 
)     Address: T. Leko, Independent Research Press, 3350 Martha Custis
)              Drive, Alexandria, VA 22302-2115 USA, Phone-Fax:
)              + 1 703 998 8093, E-mail: tleko@aol.com
) 
)     YOUR COMMENTS WILL BE GREATLY APPRECIATED !

What are these supposed to be examples of?

f(z) = 1/(1-z)^2 is -not-  analytic on the unit disc. I wasn't aware
anyone expected it to be. =I= am certainly not shocked at it. It
obviously has a pole of second order at z = 1.

The function

      f(z) = ((1-z)/(pi sqrt z)) ln((1+sqrt z)/(1-sqrt z))

is singular at z = 0. The derivative (unless I made a mistake in
calculation) is 
	               
      f'(z)= -( (2t^3-t^2+1)/(pi t^2))ln((1+t)/(1-t))+1/(pi t)

where t = z^(1/2).

As one can easily see, the only place where a singularity is possible
is where t = 0 (and hence z = 0) or t = -1 (and z = +/- i), or t = +1
(and z = +/- 1). In the first case, the limit of the derivative is 0,
but the derivative does not actually exist there. Whether this is a
removable singularity, I don't know (the work looks like low return for
effort expended to me). For t = -1, one gets a logarithmic singularity
in the derivative. Hence the function fails to be analytic on the unit
disc. For t = 1, one again gets a logarithmic singularity. So for at
least four points on the unit disc, the function is not analytic, as
expected. This is by standard analysis.

What is the path of integration supposed to be for the third example?

Frankly, I don't understand what you are getting at.

Mike
----------------------- 

Subj: Bracket a complex root
Date: 07/08/96
From: tleko@aol.com
To:   jmccarty@altair.dsccc.com

Dear Professor McCarty,

Thank you very much for your letter of July 8, 1996. 

May you please give me your mailing address to send you 
the volume seven of "Critical Comments ..' which I
presented in Prague. You will have it in a day or two
by priority mail.

Your comments will be greatly appreciated.

Best regards, Toma Leko
----------------------------------

Subj: RE: Bracket a complex root
Date: 96-07-08 17:57:39 EDT
From: JMCCARTY@altair.dsccc.com (MIKE MCCARTY -- 93789)
To: Tleko@aol.com
CC: JMCCARTY@altair.dsccc.com 

You are welcome for the letter. I am not a professor. I am a senior
level software engineer. My undergraduate degree is in mathematics, my
graduate work was in mathematical probability and statistics (mostly
measure theory on sigma algebras of sets, really). I taught at the
University of Texas at Dallas while I was a graduate student.

But I would be glad to look over whatever.

Mike
-----------------------------------------

Subj: recent correspondence
Date: 07/19/96
To: jmccarty@altair.dsccc.com

Dear Dr. McCarty,                  July 19, 1996

I read several of your posted letters and sincerely
appreciate your analytical mind. I will gladly send 
you the other volumes of 'Critical Comments ..' if
you are interested.

Best regards, tleko@aol.edu
------------------------------------------
Subj: RE: recent correspondence
Date: 96-07-22 08:40:11 EDT
From: JMCCARTY@altair.dsccc.com (MIKE MCCARTY -- 93789)
To: Tleko@aol.com
CC: JMCCARTY@altair.dsccc.com

Thank you for your comment. I would enjoy reading other volumes. I would
also appreciate it if you did not publish our correspondence. I intended
that it be kept private.

Mike
-----------------------------------

Subj: Re: How to bracket complex zeros - an example
Date: 96-07-22 09:08:48 EDT
From: JMCCARTY@altair.dsccc.com (MIKE MCCARTY -- 93789)
To: Tleko@aol.com
CC: JMCCARTY@altair.dsccc.com

I don't have access to MATLAB. But in any case, what you describe is not
what most mathematicans consider to be "braketing a root". What you
describe is more like exploring. It is useful, but not what we mean by
"braketing a root".

Mike
--------------------------------------

Dear Dr. McCarty,                                 July 22, 1996

Thank you very much for your messages today.
Aren't you looking for the zeros of a given complex polynomial ?

Peter Verthez wrote: ".. When you try to numerically find a real root of a
continuous function, you would begin by finding two x-values for which the
y-values have opposite signs. Then you know that there is a root 
{ zero value of y } in between them."

"Now take the complex case, an analytic function w=f(z) in the complex
plane (analytic in some region). Is there a way to determine a region in 
the complex plane in which you will find a complex root { zero value 
of f(z) } ?"

To show the location of the zeros of f(z)=z^2+z+1 I have generated three
plots using MATLAB pogram. For you not having the MATLAB software I will
gladly send to you the generated plots if you would give me your FAX
number.

All the comments you might have are very valuable to me and I would like
to share them with the collegues. I was very fortunate that a number of
eminent mathematicians have helped me along the way. Their work and
names are in the quoted volumes.

Sincerely yours, Toma Leko.
---------------------------------------

Subj: Re: How to bracket complex zeros
Date: 96-07-22 18:46:53 EDT
From: JMCCARTY@altair.dsccc.com (MIKE MCCARTY -- 93789)
To: Tleko@aol.com
CC: JMCCARTY@altair.dsccc.com
                                
I am not a Ph.D.

All well and good. But that is not a general technique for finding
zeroes. First, it is only as good as the resolution of the mesh you put
on the function. Second, it requires human intervention and pattern
matching. Third, it requires human insight to know -where- to look. If I
write down some arbitrary polynomial with complex coefficients, how do
you know where to go looking for roots?

Please do not publish any of our correspondence.

Mike
------------------------------------

Subj: Re: recent correspondence
Date: 07/23/96
To: jmccarty@altair.dsccc.com

Dear Dr. McCarty,

Your request for confidentiality of our correspondence does not seem to
me to serve the objective of free discussion of mathematical terms and 
does hinder the progress of mathematical science.

I regret that you may subscribe to such a policy. Please let me know.

Best regards, tleko@aol.com

----------------------------------

Subj: Re: recent correspondence
Date: 96-07-23 15:30:20 EDT
From: JMCCARTY@altair.dsccc.com (MIKE MCCARTY -- 93789)
To: Tleko@aol.com
CC: JMCCARTY@altair.dsccc.com

I am not a Ph.D.

Not only -may- I subscribe to such a policy, I -do- subscribe to such a
policy. 

Mr. Leko, I have carefully read the material you sent to me (Vol VII, I
believe). It is a very interesting compendium of information. It
carefully documents how learned men have attempted with very great
patience to explain a very elementary theorem to you, and how you have
failed completely to understand how it is that you are wrong. I was
amazed at how politely, carefully, and tactfully you were handled.

I am nervous about continuing our message exchange, lest I become
another chapter in the sad story of how you know and understand just
enough mathematics to pose this problem, but just not quite enough to
understand its solution.

You do seem to place faith in numerical computations performed by
(mechanical) computers. So, even though I am convinced that it is
hopeless, I have written up an explanation of how you can explore for
yourself, using your own numerical programs, Cauchy's theorem.

So here goes.


Consider
	    dt
	Sc ----
	   t-z

where the contour is any simple closed contour in the complex plane.
Then:
	if z is exterior to the contour, the value is 0
	if z is interior to the contour, the value is 2iPI
	if z is on the contour, the integral does not exist

Hopefully, you are willing to accept the fact that the value does not
depend on the particular contour, but only on whether z be interior to
or exterior to the contour. I will therefore restrict consideration to
circular contours.

Let the contour be a circle with radius r, centered on 0+0i. (This does
not entail any loss of generality for circles, since one can translate
the circle and z both so that the circle is centered at the origin. If,
that is, you accept that circles are sufficient.) Let z = x+iy (x and y
both real).

            [ erroneous stuff cut ]
Mike
------------------------------------------

Subj: Re: recent correspondence
Date: 07/23/96
To: jmccarty@altair.dsccc.com

Dear Dr. McCarty,

Thank you very much for your letter of July 23, 1996.

If you believe in what you wrote, I will publish it and will 
tell you where you went wrong.

Please let me know.

Best regards, tleko@aol.com

--------------------------------------------

Subj: Re: recent correspondence
Date: 96-07-23 18:14:05 EDT
From: JMCCARTY@altair.dsccc.com (MIKE MCCARTY -- 93789)
To: Tleko@aol.com
CC: JMCCARTY@altair.dsccc.com

I am not a Ph.D

Do not publish anything I write without getting prior permission.
If you discover an error in my work, please let me know.

Mike
-----------------------------------

Subj: Re: recent correspondence
Date: 07/23/96
To: jmccarty@altair.dsccc.com

Dear Dr.( I would have given you one) McCarty,

I greatly appreciate your  effort.
Yes, you have several errors.

Please try again.

Best regards,  tleko@aol.com
-----------------------------------------

Subj: Re: recent correspondence
Date: 96-07-24 10:35:25 EDT
From: JMCCARTY@altair.dsccc.com (MIKE MCCARTY -- 93789)
To: Tleko@aol.com
CC: JMCCARTY@altair.dsccc.com

In fact, there -did- creep in 5 -typographical- (not mathematical)
errors in my message to you. Three signs were reversed, and two numbers
were incorrectly typed. But the conclusions presented were valid.

If you believe that other errors (mathematical as opposed to typing
errors) please indicate where these errors lie. Simply stating "Yes, you
have several errors." is -not- helpful, to me or to you. If you have
objections, then be specific.

 Tell me where I am wrong, and I will listen to you.
Make broad statements to the effect -that- I am wrong, and I will not
listen to you

Here is a corrected version of the information I sent you. Annotations
indicate where changes were made.

Some notation (so I can use ordinary ASCII characters):

	S	integration

	Sc	integration over a closed contour
        
   @                                                             the Greek
letter Theta (sort of looks like it, I guess)

	PI	3.14159265358979...

	i	the imaginary unit, sqrt(-1)

	^2	squaring

Consider
	    dt
	Sc ----
	   t-z

where the contour is any simple closed contour in the complex plane.
Then:
	if z is exterior to the contour, the value is 0
	if z is interior to the contour, the value is 2iPI
	if z is on the contour, the integral does not exist

Hopefully, you are willing to accept the fact that the value does not
depend on the particular contour, but only on whether z be interior to
or exterior to the contour. I will therefore restrict consideration to
circular contours.

Let the contour be a circle with radius r, centered on 0+0i. (This does
not entail any loss of generality for circles, since one can translate
the circle and z both so that the circle is centered at the origin. If,
that is, you accept that circles are sufficient.) Let z = x+iy (x and y
both real).

Thus
	t = r exp(i@),

and the contour integral may be transformed into a complex integral over
a real interval

       dt     PI  ir exp(i@)d@    PI   ir(cos@+isin@)    
Sc ---- = S --------------- = S  ----------------------- d@
      t-z    -PI  r exp(i@)-z     -PI    r(cos@+isin@)-x-iy       

   PI    ircos@-rsin@                      PI      icos@-sin@  
= S  ----------------------- d@ =  S ----------------------- d@

  -PI  (rcos@-x)+i(rsin@-y)          -PI  (co@-x/r)+i(sin@-y/r)

Now, if we rescale so that x' = x/r, and y' = y/r, then rewrite using x
for x' and y for y', we see that we could have used the unit circle from
the start.

        PI    icos@-sin@
	  = S ------------------- d@
      -PI (cos@-x)+i(sin@-y)

Nothing special or funny going on.

Now I'm going to rationalize the denominator by multiplying numerator
and denominator by the complex conjugate of the denominator, then
simplify and split the integral into real and imaginary parts

   PI         (icos@-sin@)[(cos@-x)-i(sin@-y)]
= S ----------------------------------------- d@
-PI [(cos@-x)+i(sin@-y)][(cos@-x)-i(sin@-y)]

      PI icos@(cos@-x)+cos@(sin@-y)-sin@(cos@-x)+isin@(sin@-y)
	= S------------------------------------------------------ d@
	 -PI (cos@-x)^2+(sin@-y)^2

	      PI cos@(cos@-x)+sin@(sin@-y)
	= iS  ------------------------ d@
	        -PI (cos@-x)^2+(sin@-y)^2

                          PI cos@(sin@-y)-sin@(cos@-x)
                        + S -------------------------- d@
                         -PI (cos@-x)^2+(sin@-y)^2

	   PI cos^2@-xcos@+sin^2@-ysin@
	= iS  ------------------------- d@
	  -PI (cos@-x)^2+(sin@-y)^2

                          PI sin@cos@-ycos@-sin@cos@+xsin@
                        + S ------------------------------ d@
                         -PI (cos@-x)^2+(sin@-y)^2

	   PI     1-xcos@-ysin@
	= iS  --------------------- d@
	  -PI (cos@-x)^2+(sin@-y)^2

                          PI     xsin@-ycos@
                        + S ---------------------- d@
                         -PI (cos@-x)^2+(sin@-y)^2


	   PI           1-xcos@-ysin@
	= iS  ------------------------------- d@
	  -PI 1 + x^2 + y^2 - 2xcos@ - 2ysin@

                          PI           xsin@-ycos@
                        + S -------------------------------- d@
                         -PI 1 + x^2 + y^2 - 2xcos@ - 2ysin@

These two integrals may be written in terms of three ordinary real
integrals:

          PI              1
	I1 = S -------------------------------- d@
        -PI 1 + x^2 + y^2 - 2xcos@ - 2ysin@

           PI            sin@
	I2 = S -------------------------------- d@
         -PI 1 + x^2 + y^2 - 2xcos@ - 2ysin@

           PI            cos@
	I3 = S -------------------------------- d@
          -PI 1 + x^2 + y^2 - 2xcos@ - 2ysin@

Thus we see that:

	    dt	 
	Sc ---- = i(I1-xI3-yI2) + (xI2-yI3)
	   t-z            ^This sign was inadvertently
			   put as "+"

The reason I split it up that way is so I can do it on my handy little
Sharp calculator. I used Simpson's rule with 40 intervals to numerically
integrate each of those three integrals on my Sharp EL-506D calculator.
I did each one twice, once with z = 0.5 + 0.1i and once with z = 2+3i.

According to Cauchy's formula, the first should be 0+2PIi, the second
0+0i.

Here are the results:

	z = 0.5 + 0.1 i
	x = 0.5
	y = 0.1

	I1 = 8.49079055
^An extra "0" was inadvertently inserted here
	I2 = 0.849079095
	I3 = 4.245395478

	    dt	 
	Sc ---- = i[8.49079055-(0.5)(4.245395478)-
	   t-z            ^An extra "0" was inadvertently inserted here
                                                 ^This sign was reversed
  (0.1)(0.849079095)] + 
  (0.5)(0.849079095)-(0.1)(4.245395478)

	= 3e-10+6.283185307i ~ 2PIi

---------------------------------
	z = 2 + 3 i

	x = 2
	y = 3

	I1 = 0.523598775
	I2 = 0.120830486
	I3 = 0.080553657

	    dt	 
	Sc ---- = i[0.523598775-(2)(0.080553657)-
	   t-z                                  ^This sign was reversed.
                   (3)(0.120830486)] 
   (2)(0.120830486)-(3)(0.080553657)

	= 0.0000000001 + 0.00000003i ~ 0 + 0i

	--------------------------------------------

I picked those values for z "out of the air" so to speak. You may try
others yourself, and verify that Cauchy's formula works (at least to
within numerical inaccuracies).

Any other intervals for @ may be used, so long as they are 2PI in
length. You may integrate over [0,2PI], [-PI,PI], etc. The results will
vary somewhat, due to different roundoff and truncation errors, but
should reflect the fact that Cauchy's formula is correct. Just don't try
something like [1e9PI,1e9PI+2PI] or something where you will get
numerical problems with the sines and cosines being reduced in accuracy
:)
Really, Mr. Leko, I hope you go through the math I did above to verify
for yourself that I made no mistakes, and do some numerical integration
for yourself. This may alleviate your nervousness about Cauchy's
integral formula. It really does work.

Mike
-----------------------------------------

Dear Dr. McCarty,                          July 24, 1996

You did not solve any one of the integrals you mentioned in your 
letter today in terms of arbitrary x, y in the unit disk, to verify
that the real and the imaginary values of Cauchy's integral
dt/(t-z) are continuous and that the complex integral function
is analytic in the unit disk.

Thank you very much for the nummerical values at two particular 
points in the first quadrant.

Your comments will be greatly appreciated.

With best regards,  tleko@aol.com
-----------------------------------------

Subj: Re: recent correspondence
Date: 96-07-24 15:19:18 EDT
From: JMCCARTY@altair.dsccc.com (MIKE MCCARTY -- 93789)
To: Tleko@aol.com
CC: JMCCARTY@altair.dsccc.com

)Dear Dr. McCarty,
      ^^ Please

)You did not solve any one of the integrals you mentioned in your 
)letter today in terms of arbitrary x, y in the unit disk to verify
)that the real and the imaginary values of Cauchy's integral
)dt/(t-z) are continuous and that the complex integral function
)is analytic in the unit disk.

That is true. I did not. But you claimed that I made mistakes. Please
point out where the mistakes were made.

)Thank you very much for the nummerical values at two particular 
)points in the first quadrant.

You are welcome. If you will try -other- points in any of the four
quadrants, you will find that the values do -not- depend on the values
chosen for z. The value of the integral is -constant-. It is either 0
(for z outside the contour), or 2PIi (for z inside the contour). This
has already been proven a multitude of ways by others, as documented in
your Vol. VII. I see no need to reproduce that effort.

You seem to have distrust of straightforward proofs. I wanted to provide
you with a way -you- could investigate the values of the integrals
-numerically-, for purposes of verifying that the values do -not- depend
on the particular <x,y> coordinates of z. I have done that. I even did a
little computation at a couple of selected points, to show you how.

If you will actually take the time to compute (using MatLab or whatever
other numerical computation engines you trust) the value of the contour
integral using the technique I developed for you, then you will find
that the values are as predicted by Cauchy's integral formula. The
values do -not- depend on the particular point z (except insofar as it
be internal to or external to the contour). You now have at your
disposal the ability to verify this by putting in a few hundred
different values for z, and -seeing- that the value of the integral is
indeed constant.

If I am wrong, then all you have to do is plug in a few
values for z, and -show- that the integral is not constant.

For some reason, you will not do this. Why not just try it? Are you
afraid that you will be proven wrong?

Believe me when I say that I did not specially pick the values for z. I
chose them arbitrarily. If the value of the contour integral truly did
depend on the point z, then isn't it an amazing coincidence that I just
happened to pick one for which the value turned out to be 2PIi? Or am I
so smart that I can run Simpson's rule in reverse, and thus determine
where to pick my z? (And lose my integrity in doing so.)

Or do you think that the entire numerical procedure I presented is a
mistake or a sham and deception? If so, then please point out where:

	I made a mistake or misrepresentation in the mathematical
	derivation of the three integrals

	I made a mistake or misrepresentation in the numerical values
	computed using Simpson's rule on my calculator

You do not believe the mathematical approach for some reason. That's
fine. You seem to have faith in numerical approaches. I have shown you
one. If you are not familiar with numerical integration techniques,
then I can show you where to get information, or help you with them.

Perhaps the best way would be to actually numerically integrate the
complex function itself. Would that help you? If I wrote you a program
in C or BASIC or FORTRAN, which actually did numerical integration to
approximate the integral using complex functional values, would you be
willing to run the program at finer meshes for a variety of values of z,
and actually -see- whether the values depend on z?

Find a value for z (i.e. values for x and y) such that the values
computed are not as predicted by Cauchy's theorem. You show me what the
-actual- value of the integral is, either by numerical means, or by
actually performing an integration. I do not mean by antidifferen-
tiating. Antidifferentiation and integration are -not- the same thing.
Antidifferentiation is a short cut which yields the same results as
integration under certain circumstances. One of the necessary
circumstances is that no non-removable singularity is enclosed in the
contour. Another is that the contour must not be self-intersecting. In
this case, the circumstances do not obtain. Show me that the actual
value of the integral is -not- as predicted by Cauchy's theorem. Then
we will have a basis for discussion.

Mike
----------------------------------------

Dear Mike,                             July 25, 1996

You have done a great job over the weekend. Relax. 
If you really want to check Cauchy's integral you  would
have to solve in real terms (having x and y arbitrary) all
the integrals in your letter of July 24, 1996.

I will send you in a few days the other volumes of 'Critical
Comments ...'.

Best regards,  Toma

------------------------------------------

Subj: Re: recent correspondence
Date: 96-07-25 09:44:08 EDT
From: JMCCARTY@altair.dsccc.com (MIKE MCCARTY -- 93789)
To: Tleko@aol.com
CC: JMCCARTY@altair.dsccc.com

By "solve" I assume you mean "write in closed form". That has already
been done by me many years ago (like in 1975 or so) when I proved
Cauchy's residue theorem. The value of the integral is:

            dt     2PIi    for z inside the contour
	Sc --- =   0       for z outside the contour
           t-z     does not exist otherwise

The value of the integral does not otherwise depend on the value of z.
Of course, the values of I1, I2, and I3 which I showed you do
individually depend on the value of z. But taken together, they do not.
The dependencies cancel out. It's somewhat like 

	 (x-1)^2
f(x) = --------
	x^2-2x+1

One would think just to look at the formula, that the value of f depends
on the value of x. But in fact it does not.

f(x) = 1 for x ~= 1
       undefined otherwise

The numerator and denominator do, of course, depend on x. But taken
together, the ratio is constant. Except where it is undefined.

The work I did took only about 20 minutes, start to finish. It took
much longer to type in than to do. It was not "great". It was extremely
elementary calculus. The purpose of the exercise was to give you the
ability to numerically evaluate the integral for several values of z. I
have thought about writing a program for you (you pick the language, I
can write C, Pascal, BASIC, and FORTRAN) which would numerically
evaluate the integral. FORTRAN might be the one most suitable because
it directly supports a COMPLEX data type, and so the program "looks"
more like the formulas. I would put a driver around it, and let it pick
several (say 2,000) random values of z with -2 <= x <= 2 and -2 <= y <=
2, compute x^2+y^2 and see whether z lie outside or inside the unit
circle. Then compute the value of the integral numerically. Perhaps if
you saw something like that, you would begin to realize that the
integral value depends only on whether z lie inside or outside the
contour.

Have you verified the values of the integrals as I presented them, by
running MatLab or whatever your favorite numerical program is?

Have you checked my splitting of the integral into real and imaginary
parts for error?

If you find an error either in the numerical values (neglecting
roundouff and truncation errors of course) or in my manipulations of the
integrals to get the real and imaginary (the actual mathematics), please
let me know specifically where the error lies.

Mike
-----------------------------------------

Subj: Re: recent correspondence
Date: 07/25/96
To: jmccarty@altair.dsccc.com

Dear Mike,                                      July 25, 1996

Thank you for the letter. Again you gave me the
pointwise suggestion. Where is the solution in "closed
form" you mentined,  such as:

         Contour Integral   [dt/(t-x-iy)] = f(x,y,i)    ?

You got it ?  Is it analytic ?

Best regards, Toma
---------------------------------------

Subj: Re: recent correspondence
Date: 96-07-25 18:17:00 EDT
From: JMCCARTY@altair.dsccc.com (MIKE MCCARTY -- 93789)
To: Tleko@aol.com
CC: JMCCARTY@altair.dsccc.com
                           
Mr. Leko,

I have instructed you not to publish any of our correspondence. I am
especially dismayed that you would consider publishing my home address.

Mike
---------------------------------------

Subj: Re: recent correspondence
Date: 07/26/96
To: jmccarty@altair.dsccc.com

Dear Mike,                                 July 26, 1996

I deleted your home address. Would you prefer
that I mark your contribution anonymous?
Please let me know.

With many thanks,
Yours sincerely,  Toma                                                 
----------------------------------------

Dear Mike,                                      July 25, 1996
Would you like to proofread the attachment.
Best regards, Toma
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Subj: Re: recent correspondence
Date: 96-07-26 10:16:59 EDT
From: JMCCARTY@altair.dsccc.com (MIKE MCCARTY -- 93789)
To: Tleko@aol.com
CC: JMCCARTY@altair.dsccc.com

No, Toma. I believe that you are being deliberately obtuse. I have made
no contribution. I have sent private e-mail, which is copyright by me,
and may not be used without my permission. I have not given you
permission to publish what I wrote. Not any part.

Have you found any specific mistake in what I sent you? Have you tried
other points for numerical exploration?

Mike
-------------------------------------

Subj: Re: recent correspondence
Date: 07/26/96
To: jmccarty@altair.dsccc.com

Dear Mike,

Thank you very much for your letter today.
Surely your letters are your property. I am not
publishing your work. I am only quoting you.

Best regards,
Sincerely yours, Toma