已知复数z1满足:(1+2i).z1=4+3i,zn+1-zn=2+2i(n∈N+).(1)求复数z1(2)求满足|zn|≤13的最大正整数n.-数学

题目简介

已知复数z1满足:(1+2i).z1=4+3i,zn+1-zn=2+2i(n∈N+).(1)求复数z1(2)求满足|zn|≤13的最大正整数n.-数学

题目详情

已知复数z1满足:(1+2i)
.
z1
=4+3i,zn+1-zn=2+2i(n∈N+).
(1)求复数z1
(2)求满足|zn|≤13的最大正整数n.
题型:解答题难度:中档来源:不详

答案

(1)设z1=a+bi(a,b∈R),则
z1
=a-bi
(1+2i)(a-bi)=4+3i
a+2b+(2a-b)i=4+3i
a+2b=4
2a-b=3

解得:
a=2
b=1

∴z1=2+i
(2)由zn+1-zn=2+2i(n∈N*)得:
z2-z1=2+2i
z3-z2=2+2i
z4-z3=2+2i

zn-zn-1=2+2i(n∈z,n≥2)
累加得zn-z1=2(n-1)+(n-1)i(n∈N*)
zn=2n+(2n-1)i(n∈N*)
|zn|=
4n2+(2n-1)2
=
8n2-4n+1

令|zn|≤13,即8n2-4n+1≤169
2n2-n-42≤0
1-
1+8×42
4
≤n≤
1+
1+8×42
4
<5

∴n的最大整数取值是4.

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