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author | Holden Rohrer <hr@hrhr.dev> | 2021-05-05 19:03:09 -0400 |
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committer | Holden Rohrer <hr@hrhr.dev> | 2021-05-05 19:03:09 -0400 |
commit | 61b48fd8baae321daf294ebf670ef4906240d260 (patch) | |
tree | 76466290927faca6f32ace2ca79c6e079d935208 /kang/att3.tex | |
parent | 7b1a85146c6fb0d4b2232aa7bde1ec192b42599a (diff) |
homeworks and attendance quizzes for kang
Diffstat (limited to 'kang/att3.tex')
-rw-r--r-- | kang/att3.tex | 24 |
1 files changed, 24 insertions, 0 deletions
diff --git a/kang/att3.tex b/kang/att3.tex new file mode 100644 index 0000000..2afdb9a --- /dev/null +++ b/kang/att3.tex @@ -0,0 +1,24 @@ +\noindent 1. Over $|z| = 1,$ find the integral of $e^{2z}/z^4$ + +Given $f(z) = e^{2z},$ $\int_C {f(z) dz\over z^4} = {2\pi i\over 3!} +f^{(3)}(0) = {8\pi i\over 3}$ + +\noindent 2. Prove theorem. + +$|z-z_0|$ is, on $C_R,$ exactly $R.$ Therefore, the upper bound of the +absolute value of the nth derivative of $f$ at $z_0$ becomes +$${n! \over 2*\pi*i R^{n+1}} |\int_C f(z)dz|.$$ + +$$|\int_C f(z) dz| < \int_C |f(z)| dz \leq 2\pi RM_R,$$ +so $$|f^{(n)}(z_0)| \leq {n!M_R \over R^n}.$$ + +\noindent 3. If f is entire and bounded, it is a constant + +$|f(z)| <= M_R$, where $M_R$ is some number. + +This means the theorem from question 2 applies, and because $R$ can be +arbitrarily large, the first derivative at $z_0$ (any point in the +plane) must be bounded by $n!M_R/R \to 0$ as $R\to\infty$. This means +the function is constant. + +\bye |