Question

設(shè)復(fù)數(shù)z=

1+i

1-i

，則

C

0 8

+

C

1 8

•z+

C

2 8

•z² +

C

3 8

•z³+

C

4 8

•z⁴+

C

5 8

•z⁵+

C

6 8

•z⁶+

C

7 8

•z⁷=

 

．

Answer 1

考點：復(fù)數(shù)代數(shù)形式的混合運算

專題：數(shù)系的擴充和復(fù)數(shù)

分析：化簡可得z=i，原式=（

C

0 8

+

C

1 8

•z+

C

2 8

•z² +

C

3 8

•z³+

C

4 8

•z⁴+

C

5 8

•z⁵+

C

6 8

•z⁶+

C

7 8

•z⁷+

C

8 8

•z8）-

C

8 8

•z8，由二項式定理代值計算可得．

解答： 解：化簡可得z=

1+i

1-i

=

(1+i)2

(1-i)(1+i)

=i，
∴

C

0 8

+

C

1 8

•z+

C

2 8

•z² +

C

3 8

•z³+

C

4 8

•z⁴+

C

5 8

•z⁵+

C

6 8

•z⁶+

C

7 8

•z⁷
=（

C

0 8

+

C

1 8

•z+

C

2 8

•z² +

C

3 8

•z³+

C

4 8

•z⁴+

C

5 8

•z⁵+

C

6 8

•z⁶+

C

7 8

•z⁷+

C

8 8

•z8）-

C

8 8

•z8
=（1+z）⁸-

C

8 8

•z8=（1+i）⁸-1=（2i）⁴-1=16-1=15
故答案為：15

點評：本題考查復(fù)數(shù)的代數(shù)形式的混合運算，涉及二項式定理的應(yīng)用，屬基礎(chǔ)題．