2.4.10 证明下面u或v为调和函数，并求解析函数f(z)=u+iv:(a)u=x³-3xy²; (b)u=x²-y²+2x;(c)u=(x)/(x^2)+y^(2); (d)u=(x)/(x^2)+y^(2)-2y;(e)u=2e^xsin y; (f)v=2xy+3x;(g)v=(-y)/((x+1)^2)+y^(2); (h)v=arctan(y)/(x),x>0;(i)v=e^x(y cos y+x sin y)+x+y;(j)v=(y)/(x^2)+y^(2),f(2)=0.

(a) $u = x^3 - 3xy^2$，满足调和方程，$f(z) = z^3$。 (b) $u = x^2 - y^2 + 2x$，满足调和方程，$f(z) = z^2 + 2z$。 (c) $u = \frac{x}{x^2 + y^2}$，满足调和方程，$f(z) = \frac{1}{z}$。 (d) $u = \frac{x}{x^2 + y^2} - 2y$，满足调和方程，$f(z) = \frac{1}{z} - 2iz$。 (e) $u = 2e^x \sin y$，满足调和方程，$f(z) = 2ie^z$。 (f) $v = 2xy + 3x$，满足调和方程，$f(z) = -iz^2 - 3iz$。 (g) $v = \frac{-y}{(x+1)^2 + y^2}$，满足调和方程，$f(z) = \frac{i}{z+1}$。 (h) $v = \arctan \frac{y}{x}$，满足调和方程，$f(z) = i \ln z$。 (i) $v = e^x (y \cos y + x \sin y) + x + y$，满足调和方程，$f(z) = ze^z + iz$。 (j) $v = \frac{y}{x^2 + y^2}$，满足调和方程，$f(z) = -\frac{i}{z}$。 \[ \boxed{ \begin{array}{ll} (a) & z^3 \\ (b) & z^2 + 2z \\ \vdots \\ (j) & -\frac{i}{z} \\ \end{array} } \]