设z是虚数,w=z+1/z是实数,且-1

设z是虚数,w=z+1/z是实数,且-1
设z是虚数,w=z+1/z是实数,且-1

设z是虚数,w=z+1/z是实数,且-1
1、设z=x+yi(x、y∈R,y≠0),w=x+yi+1/(x+yi)=x+x/(x²+y²)+[y-y/(x²+y²)]i
由w是实数,得y-y/(x²+y²=0,由y≠0得x²+y²=1.
所以|z|=1.
于是w=2x,所以-1

设z=a+bi,(a,b∈R,b≠0)
w=z+1/z=a+bi+1/(a+bi)=a+a/(a^2+b^2)+[b-b/(a^2+b^2)]i∈R
b-b/(a^2+b^2)=0, b=b/(a^2+b^2),
b≠0, 1=1/(a^2+b^2), a^2+b^2=1,
1. |z|=√(a^2+b^2)=1
w=a+a/(a^2+b^2)=2...

全部展开

设z=a+bi,(a,b∈R,b≠0)
w=z+1/z=a+bi+1/(a+bi)=a+a/(a^2+b^2)+[b-b/(a^2+b^2)]i∈R
b-b/(a^2+b^2)=0, b=b/(a^2+b^2),
b≠0, 1=1/(a^2+b^2), a^2+b^2=1,
1. |z|=√(a^2+b^2)=1
w=a+a/(a^2+b^2)=2a, -1z的实部取值范围(-1/2,1)
2. 由于|z|=1，设z=cosa+isina,
u=(1-z)/(1+z)=(1-cosa-isina)/(1+cosa+isina)
=(1-cosa-isina)(1+cosa-isina)/(1+cosa+isina)(1+cosa-isina)
=[(1-isina)^2-(cosa)^2]/[(1+cosa)^2+(sina)^2]
=-2isina/2(1+cosa)=-isina/(1+cosa)
所以u是纯虚数
3. z=cosa+isina, 1/z=cosa-isina, w=2cosa
w-u^2=2cosa+(sina)^2/(1+cosa)^2
=2cosa-(1-cosa)^2=-[(cosa)^2-4cosa+1]
=-(cosa-2)^2+3
-1/2当cosa=-1/2时，w-u^2取得最小值-13/4
说明：题中-1

收起