文档介绍：复数第一章121??iiz=x+iy=-1欧拉公式Imz虚部实部Rez运算2①z?ImzImzz?Rez?Rez?221211??????????②zzz??ImImz?zIm?zRez?z??Re?zzRe?2112211221z?z21????iyx??x?iy2211③y?yixy?ixy?xx?21212211????yxxy??xxyy??i12211221????xx?yyyiyxx?iy?xyzzz?x④212121221221111????i????2222iyzxx?iy?x?yx?yzz22222222222z?x?iy⑤共轭复数????22y?iy??z?z??xiyxx共轭技巧P1页运算律3代数,几何表示??y,xiy?z?xz与平面点一一对应,与向量一一对应??k2=Argz=时,向量z和x轴正向之间的夹角θ,记作θ≠辐角当z0?±21±k=±0…3π<把位于-???zarg记作=≤π的叫做Argz辐角主值0000argz4如何寻找?例:z=1-i?4?z=i2?z=1+i4πz=-1??极坐标5????sincosi?r?z?x?iycos?rxsin?ry,:???i利用欧拉公式sin?e?cosi?i可得到rez?高次幂及n次方6zn凡是满足方程?n?z??n次方根,记作的ω值称为z的1??n??nk?i2?即???z?rer??rn第二章解析函数1极限2函数极限复变函数①???z?fDZ??W?都有对于任一与其对应注:与实际情况相比,定义域,值域变化??zf?z例??????②z?zz?zzfzlimf为极限当时以称A00z?z0??当☆zf??时,连续0??zzf?证明在每一点都连续例1????0证:z?zz?zz?fz??fz?z?0000???zfz在每一点都连续所以3导数????Czf?0'2时有例?C????C?zCz??z?ff??z?0'所以证:对有?C0?limlim??z?z0?0??zz????zfz3例证明不可导?????zz?f?z?zfzzx?iy解:令000?zz??????0?x?iyzz?zz?zz?000??0时,不存在,所以不可导。当?????????iy?x?zyfxzx?u,x,yy?iv,处可微,定理:处可导且满足u在,v在C-R?u?vvu??u?v???条件且?????f?iz?x?y?y?x?x?x???fzz不可导证明例4????????,y?xvy?z?x?iyux,fyzxu,v关于解:其中x,y可微?u?v不满足C-R条件所以在每一点都不可导1???1??x?y???Rezfz例5???????y0xv?xuxx,yf,z??Rez解:?u?v不满足C-R条件所以在每一点都不可导0?1???x?y??2例6:z?fz??????22222解:yx,y??ux?0x,vyyz??zf?x其中2x?0,2y?0?x?0,y?0C-R条件可得根据z?0处可导所以该函数在4解析????zf在若zzzf处解析。在的一个邻域内都可导,此时称00用C-R条件必须明确u,v????????????????g??ffg四则运算??zg?g?fz?gf????????g?g??gff?fzzee??☆?????zzz例:证明efz?ee???yieezyefxxz解:sin??cos?vu??????yxyeuxxxx则ycosy???eecosye,?ycossinv?x,yx??vu??iyx?z?xxC-R任一点条件处满足ysiny?esin???e??xy??vu????zxxz所以?ee?fsin?ze?ycosy?ie?i处处解析xx??练****求下列函数的导数????????23223222323:解y?iy?xyfxzz?z??x?y?iy??xyx?iyx???ixxyu?v?????22323222y?3?xyyy?x??xyxv?uxx,y,yx??3所以y?x?v?u?xyxy?22???据根C-R方程可得y?x?vu??2222y?x?3x3?y??yx????zf0z?0,其它点不存在导数。所以当时存在导数且导数为初等函数Ⅰ常数??Ⅱ指数函数yeiyexzsin?cos???????z?zzzz2ziz?zz④②①定义域??e?isine2cos2??ee?ee?ee?③2112Ⅲ对数函数???z?lne?zz叫做称满足的的对数函数,记作分类:类比nz的求法(经验)???arg?幅角主值目标:寻找??i可用:?rez?z?e?u?iv????ivuiiuiviivu?过程:ee??r?e,ere?ee?e?rez???????????k?iz???k??lnivu???ri2ln?rirgzln?argz2以所k?0,?1,?2??