第1章复变函数习题答案习题详解 下载本文

???????2?1?cos2?sin2??4sin2?2sin

22?22?arg?1?cos??isin???arctgsin?????????????arctg?arcctg??arctg?tg ??1?cos?2?2?2????????2??i???????????? ?1?cos??isin??2sin?cos???isin????2sine?2??2?2?2?? 另:1?cos??isin??1?cos???

??????????????isin????2sin2?i2sincos

222?22??22? ?2sin???sin?icos??2sin?cos2?22?2?2?????????isin????2??2sine2??i???2

另:1?cos??isin??cos0?isin0?cos??isin???cos0?cos???i?sin0?sin??

0??0??0??0????????i ??2sinsin?i2sincos?2sin??sin?icos??2sine22222?22?25)

???2

2i

?1?i解:

2i2i??1?i?2?2i???1?i

?1?i22r?1?i?2,arg?1?i??arg??1???arg1???i???????? ?1?i?2?cos????isin?????2e4

?4????4??4

?6)

?cos5??isin5??2

?cos3??isin3??32解:?cos5??isin5???ei5???2?ei10?

?cos3??isin3??3??cos??3???isin??3???3??e?3?i?3?e?9?i ?cos5??isin5??2 ??cos3??isin3??3ei10???i9??ei19??cos?19???isin?19?? e9. 将下列坐标公式写成复数的形式:

?x?x1?a11) 平移公式:?

y?y?b11?解:将方程组中的第二个方程乘以虚数单位加到第一个方程,得:

x?iy??x1?iy1???a1?ib1?

即:z?z1?A 2) 旋转公式:??x?x1cos??y1sin?

?y?x1sin??y1cos?解:将方程组中的第二个方程乘以虚数单位加到第一个方程,得:

x?iy?x1cos??ix1sin??y1sin??iy1cos??cos??x1?iy1??sin??y1?ix1? ?cos??x1?iy1??i??i?sin??y1?ix1??cos??x1?iy1??isin??x1?iy1? ??x1?iy1??cos??isin????x1?iy1?ei? ?z?z1?cos??isin???z1ei?

10. 一个复数乘以?i,它的模与辐角有何改变? 解:设z?re ?i?e?ii?

i??i?2?2 ??iz?ree

?re???i????2??

即:一个复数乘以?i,它的模不变,辐角减小

211. 证明:z1?z2证明:z1?z22?z1?z22?2z1?z2?22?,并说明其几何意义。

??。 2??z1?z2??z1?z2???z1?z2?z1?z2

? ?z1z1?z1z2?z1z2?z2z2

z1?z22??z1?z2??z1?z2???z1?z2?z1?z2

?? ?z1z1?z1z2?z1z2?z2z2 ?z1?z22?z1?z22?2z1z1?2z2z2?2z1?z2?22?

几何意义:平行四边形的两条对角线的平方和等于它的相邻两边平方和的2倍。 12. 证明下列各题: 1) 任何有理分式函数R?z??分式函数;

证明:设z?x?iy?z?x,y?,则:

P?z??u1?x,y??iv1?x,y?, Q?z??u2?x,y??iv2?x,y?

其中,u1?x,y?,u2?x,y?,v1?x,y?,v2?x,y?皆为关于x,y的实系数多项式。

P?z?可以化为X?iY的形式,其中X与Y为具有实系数的x与y的有理

Q?z?

P?z?u1?x,y??iv1?x,y??u1?iv1??u2?iv2??u1u2?v1v2??i?u2v1?u1v2?????X?iY 2222Q?z?u2?x,y??iv2?x,y?u2?v2u2?v2u1u2?v1v2u2v1?u1v2P?z???Y??Rz??X?iY , 2222u2?v2u2?v2Q?z? 其中:X?X,Y为具有实系数的关于x,y的有理分式函数。

2) 如果R?z?为1)中的有理分式函数,但具有实系数,那么Rz?X?iY; 证明:因为R?z?为具有实系数的有理分式函数,所以 Rz?R?z????????P?z??P?z?u1?x,y??iv1?x,y??u1?iv1??u2?iv2? ????22u2?v2?Q?z??Q?z?u2?x,y??iv2?x,y?2222 ??u1u2?v1v2??i?u2v1?u1v2?u?v?X?iY

其中:X?u1u2?v1v2u2v1?u1v2Y?, 2222u2?v2u2?v2nn?1???an?1z?an?0的根,那么a?ib也是它的根。 3) 如果复数a?ib是实系数方程a0z?a1znn?1???an?1z?an 证明:令f?z??a0z?a1z 因为a?ib是方程f?z??0的根,?f?a?ib??0 ? f?a?ib??0 又因为的系数为实数, ?f?a?ib??fa?ib?f?a?ib?

因此f?a?ib??0。即a?ib也是方程f?z??0的根。即实系数多项式的复根必共轭成对出现。 13. 如果z?e,证明: 1) z?n??it1?2cosnt znit证明:?z?e ?zn?11itnint?int???e??e?e??cosnt?isinnt???cosnt?isinnt??2cosnt nnitz?e?2) z?n1?2isinnt znit证明:?z?e

?zn?11itnint?int???e??e?e??cosnt?isinnt???cosnt?isinnt??2isinnt nnitz?e?14. 求下列各式的值: 1)

?3?i

3?i?2e?i?5?6解:?

5??i?5?5??5??2e6?32?cos?isin????163?16i ?66???5?2)

?3?i?5??i?6??2e???1?i?6

2e4

?i?4???2e?i?解:?1?i???1?i?3)

66?????4?2?e6i6?4?8ei3?23?3????8?cos?isin???8i

22???1

i?解:??1?e6

1i6???2n????1?e即:w0?4)

?n?0,1,2,3,4,5?

31313131?i,w1?i,w2???i,w3???i,w4??i,w5??i 22222222?1?i?13

2e3?i?4解:?1?i?

1???i???2n??3?4???1?i?1?62e6 ?n?0,1,2?

7?即:w0?2e?i?12i???7?7?????2?cos?isin?,w1?62e12?62?cos?isin?,

1212?1212???6w2?62eni5?45?5????62?cos?isin?

44??n15. 若?1?i???1?i?,试求n的值。 解:??1?i???1?i?

nn?1?i?n????????2?cos?isin???44????nn?n???2???isin?cos?

44n???1?i?n?sin16.

????????2?cos?isin???44????nn?n???2??cos?isin??

44n??n?n?n?n? ? sin??sin?0 ? ?k? ? n?4k?k?0,?1,?2,??

444431) 求方程z?8?0的所有根; 解:?z?8?0 ?z?3?8?38ei?

3?z?2ei??2n?3?n?0,1,2?

?1?i3,z1?2ei?即:z0?2ei?3??2,z2?2ei5?3?1?i3

2) 求微分方程y?8y?0的一般解。

'''3解:微分方程y?8y?0的特征方程为:r?8?0。由前题得:r0?1?i3,r1??2,r2?1?i3 '''?2x?? 微分方程y?8y?0有三个线性无关的特解:y0?e1?i3x,y1?e,y2?e?1?i3?x

'''?2x微分方程y?8y?0有三个线性实数特解:ecos3x,esin3x,e

xx''' 一般解为:y?c1e?2x?exc2cos3x?c3sin3x?c1,c2,c3?R?

17. 在平面上任意选一点z,然后在复平面上画出下列各点的位置:?z,z,?z,,,?解:

18. 已知两点z1与z2(或已知三点z1,z2,z3),问下列各点z位于何处? 1) z???11zz1z

1?z1?z2?; 2解:z位于z1与z2连线的中点。

2) z??z1??1???z2,其中?为实数; 解:z位于z1与z2连线上,其中??z?z2z1?z2。

3) z?1?z1?z2?z3?。 3解:z位于以z1,z2,z3为顶点的三角形的重心上。

19. 设z1,z2,z3三点适合条件z1?z2?z3?0,z1?z2?z3?1。证明:z1,z2,z3是内接于单位圆