yyx??∴2ze=22yxe?)exp(2z=yxe?2.设z=x+iy,计算)Re(1ze.解:∵z=x+iy∴222211yxyiyxxiyxz??????∴)sin(cos1222222yxyiyxyyxxzee?????∴2221cos)Re(22yxyeeyxx???3.求方程iz??ln2的解.解:∵lnz=2/?i∴由对数函数的定义有:Z=iiei???2sin2cos2/???∴所给方程的解为z=i4.求方程iez31??的解.解:∵)3sin3(cos231??iiez????=)3sin3(cos2??ieLn?根据指数函数的定义有:z=n2+i3/?或z=n(1+i3)四、证明题1.试证:zzzcossin22sin??.证明:根据正弦函数及余弦正数定义有:ieeziziz22sin22???222cossin2izizizeeiizezz?????ieeiziz222?????∴sin2z=2sinz·cosz2.证明:xnxxnnxxx2sin2sin21sinsin2sinsin???????.证明:令A=nxxxcos2coscos1?????B=sinx+sin2x+…sinnx∴inxxiixeeeBiA???????2122)1(121111xiizixxniexneee????????xnixixniexxnexixeni22212sin21sin2sin221sin2???????=)2sin2(cos2sin21sinxnixnxxn??∴xnxxnxxx2sin2sin21sinsin2sinsin??????第4章解析函数的积分理论一、单项选择题1.??cdz2(D),c为起点在0,终点在1+i的直线段.(A)0(B)1(C)2i(D)2(1+i)2.???1)(sinzAzdz.(A)0(B)10i?(C)i(D)123?i
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