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Complex Variable (Integral (Residue and pole (contour integral 裡有singular…

- - - - simple connected domains(都可拆解成很多個simple closed contour)
      - mutip conected domain(把有問題的點用simple closed contour 然ˊ後就可證明出大圈圈的積分=所有小圈圈的積分)
      - Cauchy integral formula
        用在contour 裡有singular points 時使用
        f is analytic everywhere inside&on a simple closed contour #
        
        Theorem: IF f is analytic at a point(since analytic 包含neighborhood 的概念) -> infinite analytic
        
        f is continuous in D(not include boundary), IF closed contour 積分=0
        then f is analytic through D
        (跟Cauchy Gausat 有非常相似的關係)
        (因為anti+無限可微得出此結論)
        
        f is analytic inside& on a circle-> 微分都是Bounded
        
        Liouville's Theorem
        f is entire&bounded ->>> f is constant for whole plane
        
        Fundamental Theorem of algebra
        
        Maximum theorem
        只要f is analytic & not constant in D
        則f 的max min 不會發生在D
        若再加constinous+ bounded region 就會衍生出max min 發生在boundary(continous theorem 4)
        
        Taylor series
        if f is analytic at zo , then there exists a power series at zo such that power seires = f(z)
        proof:拆成有限項->integral formula ->upper bound(and N-> infinite)>平移
        想法:::從integral formula 的1/s-z 想辦法展開成(z^n)/s^n+1的整項+某項
        收斂半徑::如果是用unique的方式想辦法展開，就是以後來套用的那個form做原本收斂半徑的變化
        但其實可以直接用singular 的概念想(也就是後面的radius of convergence)
        
        Table(主要就是e^z 跟 geometirc series，其他都是以此延伸推導)
        
        negative power
        如果f(z) 在z0 not analytic, 可以用硬展開的方式去得到series(收斂半徑要調整，要去除該點)
        而為何可以這樣做是因為Laurent's Theorem
        
        用table 的series 硬展開
        若有被弄到分母的就把原本收斂半徑去掉那點就可以了
        
        Laurent Series
        Laurent's TheoremIf f(z) is analytic on a domain(被小圓大園包住)，且仍然有uniqueness
        Laurent's Theorem是廣義的taylo series(when R1->0)
        Laurent's Theorem重要性不在級數本身，在於係數，因為Laurent's Theorem的係數是contour積分，可以藉由求得係數來得到積分值, b1 is called residue
        Special case:: when R1->0 Using Cauchy Goursat can let all the bn->0
        
        cn = an (n>=0) = b(-n) (n<=-1)
        
        藉由Laurent 的幫助
        f 在singular point 以中心點畫出的circle內外都存在power series(內部是taylor series, 外部是laruent series)(展開方式如1/1-z 就是兩者不同提出法硬展開)
        e.g D1,D2 那個例子(詳見老師講義)
        
        Residue #
        
        series convergence
        
        ABS convergence
        if the series converge at z=z1(R1)
        then the series is absolutely converge in R1
        
        uniform convergence
        if R0 is circle of convergence
        then inside R0 , series is uniform converge in every closed diskR1 (R1<R0) 重要性:只要series 在closed disk unifrom converge then 存在一個N> Nepsolon 使得所有remiander -> 0
        
        1 more item...
        
        逐項積分
        consider S(Z) is a power series of C.O.C (circle of convergence) and consider contour C inside C.O.C , then consider a g(z) is continous on C
        積分{ g(z)*S(z) } = 逐項積分
        
        3 more items...
        
        uniqueness, 因此用各種方法得到的power seies 只要f(z) 確實analytic(存在taylor series)，就一定都相同
        e.g sinhz , coshz 都可從sin,cos得來
  - - - 在求各個residue時，因為對於z1展開，其他點通常是在整個收斂半徑內，因此其他點都只有positive power 展開，不會有residue
        
        推廣::
        若f(z) 可以寫成rational function ，那麼各自residue 就是該分式的分子
      - 但因為你如果一個一個算residue 很慢，因此有沒有辦法一次算完?有的，如果我們考慮"residue at infinity " 這種residue 會包含所有的residue 值，使得我們能夠一次算完
        
        if f is analytic at R1<|Z|<無窮，則f(z) has singularity at infinity
        
        Res(z=無窮)=1/2pii * 積分f(z) dz
        
        Residue at infinity theorem
        條件::只要順時針contour 外沒有任何一個singular point
        則contour 積分f(z)dz = 2pii Res(Z=0)[1/z^2 f(1/z)]
    - - Poles
        
        存在有限項的bm
      - removable
        
        當定義那個應該不analytic的點，會發現級數收斂的相等，而任何一個f(z)只要在某點存在power series 收斂，代表analytic，因此這是假的singular point
      - essential
        
        無窮項bm
      - Residue at pole Theorem
        
        z0 is a pole of oder m of f(z)=
        f(z) = o/(faita)(z)/(z-z0)^m
        等價敘述
        
        Res(z=z0)f(z) = (o^m-1(z0)/(m-1)!
    - - analytic function zero theorem
        (a) f has a zero of order m at z0
        b)there is a function g,which is analytc and nonzero at zo
        st.f(z) = (z-zo)^m*g(z)
      - zero of analytic function is isolated
        a)if f is analytic at zo
        b)f (z0) = 0 but is not zero in any neighbor of zo
        ->>>then f != 0 through some deleted neighbor of z0
      - if f is analytic at zo at a neighbor N0
        f(z) = 0 at all points through a line or a domainD, contains zo
        then f =0 though N0
      - If f(z) = p/q
        p,q is analytic at zo
        p(zo) != 0 and q has a zero of order m at zo
        ->>>>then p/q has a pole of order m at zo
    - - improper integral
        
        no singular at real axis
        
        usual form
        if consider p/q , pq are polynomial, q(z) has no real zero but has at least one zero above the real axis
        
        real function -> 複變函數
        在contour 裡找出pole，並計算residue(若是usual form 則可以用rational form 的residue求法)
        最後將contour 積分用upper bound 消除，得到principal value
        若是even function 則/2 就是improper integral
        
        cos sin form
        
        令f(z)e^iaz
        
        residue 用pole form 求
        
        singular at real axis
        
        contour 繞過singular point(往上或往下繞)(如果是branch cut 那你基本上就只有1種繞法)
        step1: 將contour 拆成四段，目標是要將非實數軸上那兩段數值化，並且整個四段contour = 0(if no singular) = 2pii residue(if have)
        step2:分別說明外圍那段當R-> infinity 時，內圍半徑趨近於0(通常會用到參數積分)如何數值化(注意::indented path theorem 只能用在isolated singular point(若碰到branch cut 你只能用upper bound))
        step3:處理實數軸的積分，把它集成一項(題目要的長相)(通常用到參數積分)
      - change variable of definite integral
        
        將F(cos,sin) 在[0,2pi]的積分轉換成contour of unit circle integral
        step1:另z = e^i(theta) , 故e^i(-theta) = 1/z , 並作整個integral change variable
        step2:接下來就完全是contour integral(基本上都會有simple pole)(這邊的困難點就是在如何找到unit circle 裡有幾個pole然後用residue theorem算)
    - - meromorphic
        f 在 domain 內部幾乎都analytic，除了pole
        meromorphic function
        f is analytic in a domain except finite poles
        
        argument principle
        if f is meromorphic function, and has Z zeros and P poles in D
        then in the w-plane continuous change of argument/2pi = Z-P
        
        Rouche's theorem
        a)f,g are analytic in&on simple closed C
        b)|f| > |g| at each point on C
        then f & f+g has same number of zero