復變函數(shù)與積分變換1

1、習題一解答A類1求下列復數(shù)的實部與虛部、共軛復數(shù)、模與輻角。（1）i231?；（2）i13ii1??；（3）????2i5i24i3??；（4）i4ii218??解（1）??????2i31312i32i32i32i31???????所以???????????i231Re，1322i31Im???????????2i31312i31???，13131331332i3122?????????????????，kπ2i231argi231A

2、rg?????????????????210232arctan??????kk?（2）????????i25233i321ii)(1i1i13iiiii13ii1????????????????所以23i13ii1Re?????????25i13ii1Im??????????25i23i13ii1??????????，2342523i13ii122??????????????????，kπ2i1i3i1argi1i3i1Arg?????

3、????????????????????????210235arctankkπ.（3）??????????????????42i7i262i2i2i5i24i32i5i24i3??????????13i27226i7??????所以????272i5i24i3Re??????????，????132i5i24i3Im??????????，????l3i272i5i24i3???????????5若1||1||??ba，試證：11???b

4、aba。解：0|||1||1|||112222????????????babababababa然而????????babababababa?????????11|||1|222222||||||||1bbabaabababa??????????1||||||1222????aba????0||1||122????ba即11???baba6將下列復數(shù)化成三角表示式和指數(shù)表示式。（1）i；（2）1；（3）13i；（4）??π0isincos

5、1???????；（5）i12i??；（6）????32isin3cos3isin5cos5??????解：（1）2πie2πisin2πcosi???；（2）iπeisinπcosπ1????（3）3πi2e3πisin3πcos223i2123i1????????????????????；（4）????????????2icos2sin2sin22cos2i2sin22sinsincos12????????π)(0e22sin2πi

6、sin2πcos22sin2πi????????????????????；（5）???????????????????21i212i1i12i21i12i????????4πisin4πcos2=4πie2?（6）?????????????????i19i9i103i32i5eeeeeisin3cos3isin5cos5?????????isin19cos19??7當1||?z時，求||azn?的最大值，其中n為正整數(shù)，a為復數(shù)。解：由