设z=x+yi.gz=x-yi.(x,y∈R),则|z|²+zi-gzi≤0.x²+y²－2y≤0.===>x²+(y-1)²≤1.可设x=rcost,y=1+rsint.(t,r∈R,且0≤r≤1).===>x+y=(√2)rsin[t+(π/4)]+1.===>1-√2≤x+y≤1+√2....