設(shè)z=a+bi,z-1/z+1=ci,
則z-1=(z+1)·ci
(a-1)+bi=(a+1)·ci-bc
(a-1+bc)=(ac+c-b)i
則實(shí)部a-1+bc=0;c=(1-a)/b
虛部ac+c-b=0;
聯(lián)立消去c得
(1-a^2)/b-b=0;
a^2+b^2=1;
b^2=1-a^2;
z^2-z+2=(a^2-b^2+2abi)-(a+bi)+2
=(a^2-b^2-a+2)+(2ab-b)·i
=(2a^2-1)+(2a-1)bi
∴|z^2-z+2|
=√[(2a^2-1)^2+(2a-1)^2·b^2]
=√[4a^4-4a^2+1+(4a^2-4a+1)·(1-a^2)]
=√[4a^4-4a^2+1+4a^2-4a+1-4a^4+4a^3-a^2]
=√[4a^3-a^2-4a+2]
根據(jù)a^2+b^2=1得-1≤a≤1;
令u=4a^3-a^2-4a+2,則du/da=12a^2-2a-4
當(dāng)du/da=0時(shí),a=2/3或-1/2
∴當(dāng)a=2/3時(shí),u取得最小值u=32/27-4/9-8/3+2=106/27
則|z^2-z+2|的最小值是√u=√318/9
兩部相加(a-1+bc)+(ac+c-b)
=a(1+c)