复变函数与积分变换(修订版-复旦大学)课后的习题答案 下载本文

复变函数与积分变换(修订版)课后答案(复旦大学出版社)

复变函数与积分变换

(修订版)

主编:马柏林

(复旦大学出版社)

——课后习题答案

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复变函数与积分变换(修订版)课后答案(复旦大学出版社)

习题一

1. 用复数的代数形式a+ib表示下列复数

e?iπ/4;3?5i137i?1;(2?i)(4?3i);i?1?i.

①解πe?4i?cos???π???isin??π??22?????2?4???4??i??2???2?2i ?22②解: 3?5i?3?5i??1?7i?16137i?1??1+7i??1?7i???25?25i

③解: ?2?i??4?3i??8?3?4i?6i?5?10i ④解:

13?1?i?35i?31?i=?i?2?2?2i

2.求下列各复数的实部和虚部(z=x+iy)

z?a33z?a(a?); z3;???1?i3??2??;???1?i3??2??;in. ①

:∵设z=x+iy

则z?a??x?iy??a??x?a??iy?????x?a??iy??y??a??x?a??iy?a?x?i?x?a??iy??z? ∴Re?x?a?2?y2?z?a?x2?a22?z?a????y?x?a?2?y2 Im??z?a?2xy?z?a????x?a?2?y2. ②解: 设z=x+iy

∵???????????3z3?x?iy3?x?iy2x?iy?x2?y2?2xyix?iy ∴Rez??x3?3xy2, Im?z3??3x2y?y3.?x?x2?y2??2xy2??222?y?x?y??2xy??i?x3?3xy2??3x2y?y3?i③解: ∵???1?i3?3??1?i3?3??13?2????88?????1?3???1???3?2???3???1?2?????3???3?

??? ?18?8?0i??1

∴Re???1?i3???1?i3??2?????1, Im?????0. ?2?④解:

∵?3??1?3?3???1????1?i3???3?2??3???2???1??3??3?3???i

?12?????88?8?0i??1 2 / 66

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复变函数与积分变换(修订版)课后答案(复旦大学出版社)

∴Re???1?i3????1, Im???1?i3?????0. ?2???2??⑤解: ∵in?????1?k,n?2kk??. ?1?k????i,n?2k?1 ∴当n?2k时,Re?in????1?k,Im?in??0;

当n?2k?1时,Re?in??0,Im?in????1?k.

3.求下列复数的模和共轭复数

?2?i;?3;(2?i)(3?2i);①解:?2?i?4?1?5.

?2?i??2?i

②解:?3?3

?3??3

③解:?2?i??3?2i??2?i3?2i?5?13?65.

?2?i??3?2i???2?i???3?2i???2?i???3?2i??4?7i

④解:

1?i1?i22?2?2

??1?i??1?i?1??2???2?i2 4、证明:当且仅当z?z时,z才是实数.

证明:若z?z,设z?x?iy,

则有 x?iy?x?iy,从而有?2y?i?0,即y=0 ∴z=x为实数.

若z=x,x∈?,则z?x?x. ∴z?z.

命题成立.

5、设z,w∈?,证明: z?w≤z?w

证明∵z?w2??z?w???z?w???z?w??z?w?

?z?z?z?w?w?z?w?w

?z2?zw??z?w??w2

?z2?w2?2Re?z?w? 3 / 66

1?i2. 复变函数与积分变换(修订版)课后答案(复旦大学出版社)

≤z?w?2z?w2222 ?z?w?2z?w ??z?w?2 ∴z?w≤z?w.

6、设z,w∈?,证明下列不等式.

z?w?z?2Rez?w?w z?w?z?2Rez?w?w

2222??2??2z?w?z?w?2z?w22?22?

2并给出最后一个等式的几何解释.

证明:z?w?z?2Rez?w?w在上面第五题的证明已经证明了. 下面证z?w?z?2Rez?w?w.

∵z?w??z?w???z?w???z?w?z?w

?z?z?w?w?z?w2222222????2??

2?z?2Rez?w?w.从而得证.

22??2∴z?w?z?w?2z?w?22?

3几何意义:平行四边形两对角线平方的和等于各边的平方的和. 7.将下列复数表示为指数形式或三角形式

3?5i2π2π??;i;?1;?8π(1?3i);?cos?isin?. 7i?199??①解:

3?5i?3?5i??1?7i??

7i?1?1?7i??1?7i??38?16i19?8i17i??8???e其中??π?arctan. 5025519②解:i?ei??其中??

i?e

iπ2π. 2

③解:?1?eiπ?eπi

2④解:?8π1?3i?16π???π.

3?? ∴?8π1?3i?16π?e3??2?πi3

2π2π???isin? ⑤解:?cos99?? 4 / 66

复变函数与积分变换(修订版)课后答案(复旦大学出版社)

2π2π??解:∵?cos?isin??1.

99??i?π.3i2π2π???isin??1?e9?e3 ∴?cos99??322π3

8.计算:(1)i的三次根;(2)-1的三次根;(3) 3?3i的平方根. ⑴i的三次根.

解:

3ππ??i??cos?isin??cos22??132kπ?ππ2kπ?2?isin233?k?0,1,2?

∴z1?cos

ππ315531?isin??i. z2?cosπ?isinπ???i 662266229931?i z3?cosπ?isinπ??6622⑵-1的三次根 解:

3?1??cosπ?isinπ?3?cos12kπ+π2kπ?π?isin33?k?0,1,2?

∴z1?cosπ?isinπ?1?3i

3322 z2?cosπ?isinπ??1

5513i z3?cosπ?isinπ???3322⑶3?3i的平方根.

πi?22?4 解: 3?3i=6???i?6?e??22???1π2i4

∴3?3i?14?6?e?ππ??2kπ?2kπ???4?isin4?6??cos??22?14?k?0,1?

iππ??∴z1?6??cos?isin??64?e8

88??1π

πi99??z2?6??cosπ?isinπ??64?e8.

88??14199.设z?ei2πn,n?2. 证明:1?z???zn?1?0

i?2πn证明:∵z?e

∴zn?1,即zn?1?0.

∴?z?1??1?z???zn?1??0 从而1?z?z2+??zn?1?0

又∵n≥2. ∴z≠1

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