[複變]積分問題
The function f(z) is analytic in the region |z|>3 and its modulus is bounded
by |f(z)| < M|z| in that region, for some positive real number M. Use the
Cauchy's theorem and integral bounds to prove the
\int f(z) / [(z+6)^2 (z-2i)^2] dz =0
|z|=7
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嘗試用 |f(z)| < M|z| 的條件把分子 替換成 M |z|
然後可以用Cauchy integrl formula去計算 Residue at -6 跟2i
但是現在含有|z| 項 就不知道如何算Residue了，
可是如果不替換 f(z) 在那邊也動彈不得...
還有|z|<3 需要考慮嗎? 萬一 裡面有singularity 怎辦?
這個積分還會是0?
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