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Anowar
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This problem is from Mathematical methods for physicists by Arfken, problem 6.4.7.
A function f(z) is analytic within a closed contour C (and continuous on C). If f(z) ≠ 0 within C and |f(z)|≤ M on C, show that |f(z)|≤ M for all points within C.
The hint is to consider w(z) = 1/f(z).
I have tried to solve this in this way using Gauss' mean value theorem. Suppose, there is a point within C where |f(z)|≥ M. As |f(z)|≤ M on C, there has to be a local maximum of |f(z)| within C. We call that point of local maximum z°. Then we can take a circle with a however small radius r around z°. From Cauchy's integral theorem,
f(z°) = (1/2πi)∫c[f(z)/(z-z°)] dz (here c is the small circle of radius r)
Now taking, z = z° + r*exp(iθ)
f(z°) = (1/2π)∫ f(z° + r*exp(iθ)) dθ
Taking the modulus, |f(z°)| = 1/2π |∫ f(z° + r*exp(iθ)) dθ| ≤ 1/2π ∫ |f(z° + r*exp(iθ))| dθ
(in all cases above, the integral is taken from 0 to 2π).
Now, m(f) = 1/2π ∫ |f(z° + r*exp(iθ))| dθ is the mean value of |f(z)| on the circle of radius r. From above, we rewrite,
|f(z°)| ≤ m(f).
Again, there must be at least one point z1 on the circle where |f(z1)| is larger than m(f) or the |f(z)| is equal to m(f) for all the points on the circle. So, |f(z1)| ≥ m(f) ≥ |f(z°)|. We can take r as small as we want to make z1 as nearer to z°. So, there is point z1 in the neighborhood of z° for which, |f(z1)| ≥ |f(z°)|, so |f(z°)| can't be a local maximum, and thus there is no point inside C such that |f(z)| ≥ M.
Now, my problem is that, I didn't use the condition f(z) ≠ 0 within C for this proof. So, there should be a problem with my solution, but I couldn't find any. Now, thinking about the hint, taking w(z) = 1/f(z), we don't know whether w(z) is continuous on C or not. For, f(z) = 0 for some point on the contour C, w(z) is not continuous and we can't take Cauchy integral of w(z) on C to be zero. But, for f(z), as it is continuous on C and analytic inside, it's contour integral on C would be zero from Cauchy-Goursat theorem. I think I could use this to solve the problem, but I failed to find any path to do so. I would appreciate it very much if someone can point out where I am doing wrong or give some more hints to solve this. I am sorry for the bad format of equations as I am not yet used to using latex.
A function f(z) is analytic within a closed contour C (and continuous on C). If f(z) ≠ 0 within C and |f(z)|≤ M on C, show that |f(z)|≤ M for all points within C.
The hint is to consider w(z) = 1/f(z).
I have tried to solve this in this way using Gauss' mean value theorem. Suppose, there is a point within C where |f(z)|≥ M. As |f(z)|≤ M on C, there has to be a local maximum of |f(z)| within C. We call that point of local maximum z°. Then we can take a circle with a however small radius r around z°. From Cauchy's integral theorem,
f(z°) = (1/2πi)∫c[f(z)/(z-z°)] dz (here c is the small circle of radius r)
Now taking, z = z° + r*exp(iθ)
f(z°) = (1/2π)∫ f(z° + r*exp(iθ)) dθ
Taking the modulus, |f(z°)| = 1/2π |∫ f(z° + r*exp(iθ)) dθ| ≤ 1/2π ∫ |f(z° + r*exp(iθ))| dθ
(in all cases above, the integral is taken from 0 to 2π).
Now, m(f) = 1/2π ∫ |f(z° + r*exp(iθ))| dθ is the mean value of |f(z)| on the circle of radius r. From above, we rewrite,
|f(z°)| ≤ m(f).
Again, there must be at least one point z1 on the circle where |f(z1)| is larger than m(f) or the |f(z)| is equal to m(f) for all the points on the circle. So, |f(z1)| ≥ m(f) ≥ |f(z°)|. We can take r as small as we want to make z1 as nearer to z°. So, there is point z1 in the neighborhood of z° for which, |f(z1)| ≥ |f(z°)|, so |f(z°)| can't be a local maximum, and thus there is no point inside C such that |f(z)| ≥ M.
Now, my problem is that, I didn't use the condition f(z) ≠ 0 within C for this proof. So, there should be a problem with my solution, but I couldn't find any. Now, thinking about the hint, taking w(z) = 1/f(z), we don't know whether w(z) is continuous on C or not. For, f(z) = 0 for some point on the contour C, w(z) is not continuous and we can't take Cauchy integral of w(z) on C to be zero. But, for f(z), as it is continuous on C and analytic inside, it's contour integral on C would be zero from Cauchy-Goursat theorem. I think I could use this to solve the problem, but I failed to find any path to do so. I would appreciate it very much if someone can point out where I am doing wrong or give some more hints to solve this. I am sorry for the bad format of equations as I am not yet used to using latex.
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