已知u+v=(x—y)(x2+4xy+y2)一2(x+y) 试确定解析函数f(z)=u+iv．请帮忙

正确答案：把所给式子对xy求偏导有 ①式+②式得 u=3x²—3y²一2 ③ ①式一②式得 u=12xy． ④ 由④式对x积分有 v=∫12xydx=6x²y+φ(y) ⑤ 对⑤式两端对y微分有v_y=6x²+φ'(y) 则由u_x=v_y有3x²一3y²一2=6x²+φ'(y)即 φ'(y)=3x²一3y²一2—6x²=一3x²一3y²一2． 故φ(y)=∫(一3x²一3y²一2)dy=一3x²y—y³—2y+c 所以代入⑤式有v=3x²y—y³一2y+c． 则u=(x—y)(x²+4xy+y²)一2(x+y)一(3x²y—y³一2y+c) =x³一3xy²一2x—c 故f(z)=u+iv=(x³一3xy²一2x+c)+i(3x²y—y³—2y+c) =(x+iy)²一2(x+iy)+k =z³一2z+k(k为任意常数)．
把所给式子对x,y求偏导,有①式+②式得u=3x2—3y2一2,③①式一②式得u=12xy．④由④式对x积分,有v=∫12xydx=6x2y+φ(y),⑤对⑤式两端对y微分,有vy=6x2+φ'(y)则由ux=vy,有3x2一3y2一2=6x2+φ'(y),即φ'(y)=3x2一3y2一2—6x2=一3x2一3y2一2．故φ(y)=∫(一3x2一3y2一2)dy=一3x2y—y3—2y+c,所以代入⑤式,有v=3x2y—y3一2y+c．则u=(x—y)(x2+4xy+y2)一2(x+y)一(3x2y—y3一2y+c)=x3一3xy2一2x—c,故f(z)=u+iv=(x3一3xy2一2x+c)+i(3x2y—y3—2y+c)=(x+iy)2一2(x+iy)+k=z3一2z+k(k为任意常数)．