习题一解答1．求下列复数的实部与虚部、共轭复数、模与辐角。（1）i231+；（2）i13ii1−−；（3）()()2i5i24i3−+；（4）i4ii218+−解（1）()()()2i31312i32i32i32i31−=−+−=+所以133=⎭⎬⎫⎩⎨⎧+i231Re，1322i31Im−=⎭⎬⎫⎩⎨⎧+,()2i31312i31+=+，13131331332i3122=⎟⎠⎞⎜⎝⎛−+⎟⎠⎞⎜⎝⎛=+，kπ2i231argi231Arg+⎟⎠⎞⎜⎝⎛+=⎟⎠⎞⎜⎝⎛+�,2,1,0,232arctan±±=+−=kkπ（2）()()()()i,25233i321ii)(1i1i13iiiii13ii1−=+−−−=+−+−−−=−−所以,23i13ii1Re=⎭⎬⎫⎩⎨⎧−−25i13ii1Im−=⎭⎬⎫⎩⎨⎧−−25i23i13ii1+=⎟⎠⎞⎜⎝⎛−−，2342523i13ii122=⎟⎠⎞⎜⎝⎛−+⎟⎠⎞⎜⎝⎛=−−，kπ2i1i3i1argi1i3i1Arg+⎟⎠⎞⎜⎝⎛−−=⎟⎠⎞⎜⎝⎛−−�,±,±,=,+−=210235arctankkπ.（3）()()()()()()()()()42i7i262i2i2i5i24i32i5i24i3−−=−−−+=−+13i27226i7−−=−−=所以()()272i5i24i3Re−=⎭⎬⎫⎩⎨⎧−+，()()132i5i24i3Im−=⎭⎬⎫⎩⎨⎧−+，1()()l3i272i5i24i3+−=⎥⎦⎤⎢⎣⎡−+()()22952i5i24i3=−+，()()()()kππkπ2726arctan22i2i52i43argi2i52i43Arg+−=+⎥⎦⎤⎢⎣⎡−+=⎥⎦⎤⎢⎣⎡−+()�,2,1,0,12726arctan±±=−+=kkπ.（4）()()()()ii141iii4ii4ii10410242218+−−−=+−=+−3i1i4i1−=+−=所以{}{}3i4iiIm1,i4iiRe218218−=+−=+−3i1i4ii218+=⎟⎠⎞⎜⎝⎛+−，10|i4ii|218=+−()()()2kπ3i1arg2kπi4iiargi4iiArg218218+−=++−=+−=.2,1,0,k2kπarctan3�±±=+−2．如果等式()i13i53yi1x+=+−++成立，试求实数x,y为何值。解：由于()()[]()()()3i53i53i53yi1x3i53yi1x−+−−++=+−++()()()()[]343y51x3i3y31x5−++−+−++=[]()i1185y3xi43y5x341+=−+−+−+=比较等式两端的实、虚部，得⎩⎨⎧=−+−=−+34185334435yxyx或⎩⎨⎧=+−=+52533835yxyx解得11,1==yx。3．证明虚单位i有这样的性质：-i=i-1=i。4．证明21)||116)Re()(),Im()()22izzzzzzzz=z=+=�−2证明：可设izxy=+，然后代入逐项验证。5．对任何，是否成立？如果是，就给出证明。如果不是，对那些值才成立？z2||zz=222z解：设，则要使成立有izxy=+2||zz=2222ixyxyx−+=+y0，即。由此可得为实数。2222,xyxyxy−=+=z6．当时，求的最大值，其中n为正整数，a为复数。1||≤z||azn+解：由于|a||a||z|aznn+≤+≤+1，且当naezargi=时，有()|a|ea|a|eea|zaannan+=+=+⎟⎟⎠⎞⎜⎜⎝⎛=+|11argiargiargi故为所求。||1a+8．将下列...
