复数和数列问题求助
第一题:若复数z满足(1+i)z+(1-i)z共轭复数=2,则z模的最小值= 第二题:无穷等比数列{a*n}中,公比为q(q不等于-1),已知a*1+a*2=3(a*3+a*4),a*5=1,则lim(a*1+a*3+...+a*(2n-1))=
z=x+yi--->z~=x-yi (1+i)z+(1-i)z~=2 --->(1+i)(x+yi)+(1-i)(x-yi)=2 --->(x-y)+i(x+y)+(x-y)-i(x+y)=2 --->x-y=1--->y=x-1 --->|z|^2=x^2+y^2 =x^2+(x-1)^2 =2x^2-2x+1 =2(x-1/2)^2+1/2 所以x=1/2,y=-1/2--->z=(1-i)/2时,|z|min=1/√2。
2)等比数列{an}中,公比q<>-1。a1+a2==3(a3+a4) --->a1+a1q=3(a1q^2+a1q^3) --->a1(1+q)=3a1*q^2(1+q) q<>0,-1,a1<>0--->q^2=1/3 a5=1--->a1q^4=1 & q^2=1/3---->a1=1/q^4=9。
n-->+,lim(a1+a3+a5+。。。。。。)=a1/(1-q^2)=9/(1-1/3)=27/2。
1. z=a+bi ===> (1+i)z+(1-i)z共轭复数 = (1+i)(a+bi)+(1-i)(a-bi) = [(a-b)+(a+b)i] +[(a-b)-(a+b)i] = 2(a-b) = 2 ==> a-b = 1 z模 = 根号(a^2+b^2) = 根号[2*(b -1/2)^2 +1/2] >= 根号[1/2] ==> z模的最小值 = (根号2)/2 2. a*1+a*2=3(a*3+a*4) ===> a +aq = 3*(a*q^2 + a*q^3) ==> q^2 = 1/3 a*5=1 ===> a*q^4 = 1 ==> a = 9 因此,lim(a*1+a*3+...+a*(2n-1)) = a/(1 - q^2) = 9/(1 -1/3) = 27/2
a_1(1+q)=3a_3(1+q) 所以a_1=3a_1q^2 得q^2=1/3 又由于a_5=1=a_1q^4=a_11/9 得到a_1=9 a_1+a_3+a_5+...+a_(2n-1)=(9(1-(1/3)^n))/(1-(1/3))=...
答:z=(1-3i)/(1+i)=-1-2i 所以z位于第三象限详情>>