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§7.0FundamentalACCorrespondingtoSection7.1‐7.4in701ReviewofComplex702AC703Phasor704Impedance&AdmittanceIntroductiontoElectronic 701ComplexComplexnumberinrectangularcoordinatej -
z=a+jb Re(z)=babaaIntroductiontoElectronic Additionofcomplexz1=a+jbz2=c+jdz1+z2=a+jb+c+jd=a+c+j(b+ThisisvectorIntroductiontoElectronic Numerical543210
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- - - - IntroductiontoElectronic IntroductiontoElectronic Multiplicationofcomplexz1=a+jbz2=cz1z2=(a+jb)(c+jd)=ac+j(ad+bc)j2=-z1z2=ac-bd+IntroductiontoElectronic Numerical (2+2j)(2j)=(-4z1z142r1= 22r2= 0 - - - - IntroductiontoElectronic 小結(jié):復(fù)數(shù)的四則相等 和差 積 商復(fù)數(shù)運(yùn)算的交換律、結(jié)合律和分配在復(fù)數(shù)域中,復(fù)數(shù)沒有大小IntroductiontoElectronic Rectangular
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+j(-1/3!3 sinIntroductiontoElectronic 討復(fù)數(shù)的幅角不能唯一地確定.如果0是其中一個(gè)幅角,則 值幅角,記為argz.當(dāng)r=1時(shí),z=cosisin= 稱為單位復(fù)數(shù)IntroductiontoElectronic ConvertingfrompolartoCartesianformConvertingfrompolartoCartesianformofrepresentationofcomplexnumbers5a=3cos(/3)=b=3sin(/3)=z=1.5+
43210 - - - - IntroductiontoElectronic ConvertingfromCartesiantopolar54z=1+41212r 4.120=tan-1(4)=(=76oz=4.12e
--2 - - - - IntroductiontoElectronic ExpressthefollowingExpressthefollowingnumbersincomplexexponentialnotationZ=√3-j2=?Z=3=?Z=-1=?Z=-4j=?Z=j=?Z=cos+jsin=IntroductiontoElectronic z1=j=z2=j2=ej-
Important3z=j3=ej3-30z4=j4=ej2-- IntroductiontoElectronic Multiplicationofcomplexnumbersissimplerinz1=a+jb= z2=c+Multiplicationofcomplexnumbersissimplerinz1z2=ac-bd+z1z2 r1r2ej(Theanglebehaveslikeantan=b/atan=
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43210 - - - - -IntroductiontoElectronic AbsolutevaluesquaredofcomplexnumberisAbsolutevaluesquaredofcomplexnumberisequaltothenumbertimesitscomplexconjugatecomplex(a+jb)=(a- (a-jb)=z =(a+jb)(a-z =zz =(a+jb)(a-z =zIntroductiontoElectronic 極坐標(biāo)下的共軛zx+iycosisinrei的共軛復(fù)數(shù)為IntroductiontoElectronic Moreaboutcomplexzz=az*=a-
5 3zz=rej10z*=re- -0-Re{z}=(z+z*)/Im{z}=(z-z*)/
-- -- - - - - z =zz*=rejre-jrIntroductiontoElectronic 復(fù)數(shù)的乘冪當(dāng)r=1
k=0,1,2,…, IntroductiontoElectronic 剛才復(fù)習(xí)了復(fù)數(shù)(complex(complex IntroductiontoElectronic 補(bǔ)充概念:復(fù)變?cè)O(shè)Dz=x+iy把復(fù)變函數(shù)f(z)的實(shí)部和虛部分別記作u(x,y)v(x,y),即f(z)=u(x,y)+iv(x,y).這就是說(shuō),復(fù)變函數(shù)可以歸結(jié)為IntroductiontoElectronic 補(bǔ)充:復(fù)變函數(shù)的z+△z∈D,如果極 顯然,函數(shù)必須在點(diǎn)z連續(xù),才有可能在z點(diǎn)可導(dǎo)IntroductiontoElectronic IntroductiontoElectronic 二、復(fù)變函數(shù)可微的即若記△z=△x+i△y,由于△z=△x+i△y,無(wú)論按何方式趨于零,上式總成IntroductiontoElectronic 先看△z沿實(shí)軸趨于零的情況,此時(shí)△y≡0,△z=△x→0,以再讓△z沿虛軸趨于零,此時(shí)△x≡0,△z=i△y→0,所IntroductiontoElectronic ,定理:f(z)=u(x,y)+iv(x,y)在z=x+iyu(x,y),證明從略IntroductiontoElectronic 三、解析 區(qū)域上解析 IntroductiontoElectronic 補(bǔ)充:復(fù)變函數(shù)的冪級(jí)表達(dá)其中從收斂性如果極限存 =S,則稱復(fù)級(jí)數(shù)收斂 收斂,則 絕對(duì)收斂收斂性的比值判別
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1,IntroductiontoElectronic 若在區(qū)域G上,對(duì)任給的小正數(shù),存在一個(gè)與z無(wú)關(guān) kn(z)其中p為任意正整數(shù),則稱復(fù)變k1在G上一致收斂IntroductiontoElectronic 二、冪z0和各個(gè)ak都是復(fù)常數(shù),ak為系數(shù),z0為冪級(jí)數(shù)的中心絕對(duì)收斂
|z‐z0|<R|z‐z0|>R,發(fā)散IntroductiontoElectronic 其余的z(|z|=1)具體而定。
發(fā)散,由 IntroductiontoElectronic 羅朗解析部分:羅朗級(jí)數(shù)中n=0,∞的部分主要部分:羅朗級(jí)數(shù)中n=-1,-∞的部分羅朗級(jí)數(shù)的主要部分作變換=1/(z‐b)后也是泰勒數(shù),設(shè)新的泰勒級(jí)數(shù)收斂半徑是1R2,IntroductiontoElectronic 三、泰勒設(shè)f(z)在圓|z‐a|<R(0<R<+∞)內(nèi)解析,則f(z)在此圓內(nèi)nezn
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