就係依條的part(a。
點證，一一映射的解析函數，其一階導數不為0？
ThomasYoung
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就係依條的part(a。
反證
我前幾日睇完咗estimation lemma，距離complex analysis邁進了一步
具體點證？我試左幾日，做來做去都做唔到。
問AI
唔識rigorious prove ，但個思路應該係assume there exist z0 such that f'(z0) = 0. 而conplex function d 到一次就可以d無限次(又或者用題目 f 係 analytic). f'(z0) = 0 = f''(z0) = ... = f^無限次(z0). 即係話響z0 附近嘅點(i.e. w) , f(z0) = f(w) , but z0 =/= w. contradict to f is one to one.
好似可以用Taylor Series Expansion去寫 f(w1) = 得第一個term of Taylor Series Expansion keep 到 = f(z0)
Assume f’(z)=0
f(z) = c for any constant C.
So f(z) is non a one to one function which is contraindicated.
.’. f’(z)=/=0 🙏🏻
f(z) = c for any constant C.
So f(z) is non a one to one function which is contraindicated.
.’. f’(z)=/=0 🙏🏻
Part b
If g(w)=f-1(w)
So f(g(w))=w
differentiate both wrt w
f’(g(w))•g’(w)=1
g’(w)= 1/(f’(g(w))
If g(w)=f-1(w)
So f(g(w))=w
differentiate both wrt w
f’(g(w))•g’(w)=1
g’(w)= 1/(f’(g(w))
上邊冇寫“解”，全部零分
點解你地可以甘勁，我睇唔明
太耐無做，我諗應該係咁
v唔係constant會唔會影響m distinct roots
應該要argue for small enough |w| ? (仲細過epsilon)
應該要argue for small enough |w| ? (仲細過epsilon)
有個圖像啲嘅諗法，未必係答題會用
照例WLOG, f(0)=f'(0)=0
=> f(z) = ak z^k + ak+1 z^(k+1) + ... for some integer k >= 2
For small enough r, the z^k term dominates such that f maps the circular path
{ r e^(i theta) : 0 <= theta < 2pi }
to a loop going around the origin k >= 2 times, as theta goes from 0 to 2pi
=> Such image loop must intersect itself
=> Contradict with f being one-to-one
照例WLOG, f(0)=f'(0)=0
=> f(z) = ak z^k + ak+1 z^(k+1) + ... for some integer k >= 2
For small enough r, the z^k term dominates such that f maps the circular path
{ r e^(i theta) : 0 <= theta < 2pi }
to a loop going around the origin k >= 2 times, as theta goes from 0 to 2pi
=> Such image loop must intersect itself
=> Contradict with f being one-to-one
你可以喺個disc 入面fix 一個w_0.
只要v(w_0) 唔係0, 咁w_0 就有m個distinct roots, 每個root 都只係差個角度, 所以仲喺個disc 入面
只要v(w_0) 唔係0, 咁w_0 就有m個distinct roots, 每個root 都只係差個角度, 所以仲喺個disc 入面