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There is a recently posted Complex Analysis problem on the topics of
finding
zeros  of the equation

                       e^z-2z-1 = 0 ,   z= x+iy .


Subject: Complex Analysis problem
From: jdenny@math.fsu.edu
Date: 10 Aug 1996 15:24:44 GMT
Message-ID: <4ui9js$mo6@news.fsu.edu>

I have been unable to solve the following problem that seems to be
a typical Rouche's theorem type problem:

Show that there is exactly one solution of e^z = 2z + 1 in the unit
disk.

Thanks for the help!

                                Jeff Denny

                              Graduate Student

              Florida State University Mathematics Department

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                      The ensuing correspondence

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Subject: Re: Complex Analysis problem
From: Dmitry Gokhman <gokhman@zakuski.math.utsa.edu>
Date: Sun, 11 Aug 1996 21:26:50 -0500
Message-ID: <320E966A.6983@zakuski.math.utsa.edu>

jdenny@math.fsu.edu wrote:
> I have been unable to solve the following problem that seems to be
> a typical Rouche's theorem type problem:
> Show that there is exactly one solution of e^z = 2z + 1 in the unit
> disk.

Jeff: Let z=x+iy.
Then |e^z|=e^x and |z|=1 => |2z+1|=sqrt((2x+1)^2+(2y)^2)=
sqrt(4x^2+4x+1+4y^2)=sqrt(4x+5).
We need e^x < sqrt(4x+5) on [-1,1].
One way to see this is to observe that e^x is concave up,
whereas sqrt(4x+5) is concave down.  Thus sqrt(4x+5)-e^x
is concave up.  On the other hand, it is positive at
the two endpoints: -1 and 1.  - D

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Subject: Re: Complex Analysis problem
From: Dmitry Gokhman <gokhman@zakuski.math.utsa.edu>
Date: Tue, 13 Aug 1996 17:19:23 -0500
Message-ID: <3210FF6B.61D2@zakuski.math.utsa.edu>

Oops.  I meant to say:
Thus sqrt(4x+5)-e^x is concave *down*.
 
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Subject: Re: Complex Analysis problem
From: israel@math.ubc.ca (Robert Israel)
Date: 12 Aug 1996 00:57:43 GMT
Message-ID: <4ulvi7$o9r@nntp.ucs.ubc.ca>

In article <4ui9js$mo6@news.fsu.edu>,  <jdenny@math.fsu.edu> wrote:

>Show that there is exactly one solution of e^z = 2z + 1 in the unit
>disk.

Hint: yes, it is a typical Rouche's Theorem question.  Approximate 
f(z) = e^z - 2 z - 1 by a linear approximation g(z) = -z, and show
that |f(z) - g(z)| < 1 = |g(z)| for z in the unit circle.

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Subject: Re: Complex Analysis problem
From:  tleko@aol.com (Tleko)
Date:  21 Aug 1996 22:22:05 -0400
Message-ID: <vgg8d$k74@newsbf02.news.aol.com>

In article <4ui9js$mo6@news.fsu.edu>,  <jdenny@math.fsu.edu> wrote:

>Show that there is exactly one solution of e^z = 2z + 1 in the unit
>disk.

In article  <4ulvi7$o9r@nntp.ucs.ubc.ca>,  <israel@math.ubc.ca> wrote:

>Hint: yes, it is a typical Rouche's Theorem question.  Approximate 
>f(z) = e^z - 2 z - 1 by a linear approximation g(z) = -z, and show
>that |f(z) - g(z)| < 1 = |g(z)| for z in the unit circle.

Rouche's Theorem requires the functions to be continuous in the unit
circle.
Function z = x+iy  is however not continuous in the unit circle. We have

                       z = sqrt(x^2+y^2) * exp(i*arctan(y/x))

where arctan is not continuous in the unit circle. 
                  
Here is a MATLAB program to solve the equation

                         e^z-2z-1 = 0 ,     z=x+iy.

We have for the real part (plot 1)

                  exp(x) cos(y) - (2x+1)

[x,y]=meshdom(-1:.05:1,-1:.05:1);
colormap([1 1 1]);
z=sqrt((exp(x).*cos(y)-(2*x+1)).^2+(exp(x).*sin(y)-2*y).^2).*cos(atan((exp
(x).*sin(y)-2*y)./(exp(x).*cos(y)-(2*x+1))));
surf(x,y,z)

and for the imaginary (plot 2)

                   exp(x) sin(y) - 2y .

[x,y]=meshdom(-1:.05:1,-1:.05:1);
colormap([1 1 1]);
z=sqrt((exp(x).*cos(y)-(2*x+1)).^2+(exp(x).*sin(y)-2*y).^2).*sin(atan((exp
(x).*sin(y)-2*y)./(exp(x).*cos(y)-(2*x+1))));
surf(x,y,z)

Taking the difference of the two we search the zeros identity by using
MATLAB's contour control command (plot 3)

x=-1:.01:1;
y=-1:.01:1;
[X,Y]=meshgrid(x,y);
Z=(sqrt((exp(X).*cos(Y)-(2*X+1)).^2+(exp(X).*sin(Y)-2*Y).^2).*(sin(atan((e
xp(X).*sin(Y)-2*Y)./(exp(X).*cos(Y)-(2*X+1))))-cos(atan((exp(X).*sin(Y)-2*Y
)./(exp(X).*cos(Y)-(2*X+1))))));
contour(x,y,Z)

Plot 3 shows a  zero  solution  for x=y=0.

Using Professor Habetha's shift formula we may write 

     cos(Y+pi) = -cos(Y) ,     sin(Y+pi) = -sin(Y) ,

to obtain another solution 

     X = -.7388 ,     Y = 0 .

in plot 4.

tleko@aol.com

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Subject: Re: Complex Analysis problem
From: bruck@pacificnet.net (Ronald Bruck)
Date: Wed, 21 Aug 1996 21:38:11 -0700
Message-ID: <bruck-2108962138110001@max-14.pacificnet.net>

In article <4vgg8d$k74@newsbf02.news.aol.com>, tleko@aol.com (Tleko)
wrote:

:In article <4ui9js$mo6@news.fsu.edu>,  <jdenny@math.fsu.edu> wrote:
:
:>Show that there is exactly one solution of e^z = 2z + 1 in the unit
:>disk.
:
:In article  <4ulvi7$o9r@nntp.ucs.ubc.ca>,  <israel@math.ubc.ca> wrote:
:
:>Hint: yes, it is a typical Rouche's Theorem question.  Approximate 
:>f(z) = e^z - 2 z - 1 by a linear approximation g(z) = -z, and show
:>that |f(z) - g(z)| < 1 = |g(z)| for z in the unit circle.

:In article <4vgg8d$k74@newsbf02.news.aol.com>, tleko@aol.com (Tleko)
wrote:

:>Rouche's Theorem requires the functions to be continuous in the unit
:>circle.
:>Function z = x+iy  is however not continuous in the unit circle. We have
:
:>                 z = sqrt(x^2+y^2) * exp(i*arctan(y/x))
:
:>where arctan is not continuous in the unit circle.

Well, sci.math has just hit a new low.  The identity map is not
continuous!

--Ron Bruck
  
Yes,  you are absolutely right. There are exactly two solutions.

Best regards, tleko@aol.com

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