Question

不定积分法适用于所有求解析函数吗?u=x^2-y^2+xy,f(i)=-1+i用不定积分怎么求.

Answer 1

根据全微分公式和Cauchy-Riemann方程,可得
dv=v_xdx+v_ydy=-u_ydx+u_xdy
=(2y-x)dx+(2x+y)dy=2(xdy+ydx)-xdx+ydy
=2d(xy)-d(x^2/2)+d(y^2/2)=d({y^2-x^2}/2+2xy),
故,v={y^2-x^2}/2+2xy+C,
所以,f(z)=u+iv=(x^2-y^2+xy)+i({y^2-x^2}/2+2xy+C),
再根据f(i)=-1+i, 可得C=1/2,
故,f(z)=u+iv=(x^2-y^2+xy)+i({y^2-x^2}/2+2xy+1/2)
=(x^2-y^2+2xyi)-i(x^2-y^2+2xyi)/2+i/2
=(x+yi)^2-i(x+yi)^2/2+i/2
=z^2-iz^2/2+i/2=(1-i/2)z^2+i/2