Question

求解一道复变函数题：求解析函数f(z)，使z的虚部v(x.y)=2x^2+x-2y^2，并且f(1)=3i

Answer 1

v(x.y)=2x^2+x-2y^2

偏u/偏x=偏v/偏y=-4y

偏u/偏y=-偏v/偏x=-4x-1

u=-4xy+g(y)

-4x+f'(y)=-4x-1

g'(y)=-1

g(y)=-y+C1

f(x+iy)=-4xy-y+C1+i[2x^2+x-2y^2+C2]

∵f(1)=3i

∴C1=0 2+1+C2=3 C2=0

∴

f(x+iy)=-4xy-y+i(2x^2+x-2y^2)