Let, $u(x,y)$ be thereal part of an entire function $f(z)=u(x,y)+iv(x,y)$ for $z=x+iy\in \mathbb C$. If $C$ is the positively oriented boundary of a rectangular region $R$ in $\mathbb R^{2}$ then the value of $\int_{C} \bigl[\dfrac{\partial u}{\partial y}dx-\dfrac{\partial u}{\partial x}dy\bigr]=?$
(a) 1
(b) 0
(c) $\pi$
(d) $2\pi$
Since the function is entire , so using the Cauchy-Riemann equation we have, $\int_{C}-dv.$ But from here I can't proceed further.
since the interior of the region of integration contains no singularities, the contour integral of an exact differential gives zero, so you have answered the question. it may be useful to set out the process in logical steps as follows.
for $f$ entire $$ f(z)=f(x+iy)=u(x,y)+iv(x,y) $$ so $$ u_x+iv_x =\partial_x f=f'\partial_xz=f' $$ and $$ u_y+iv_y =\partial_y f=f'\partial_yz=if' $$ multiplying the second eqn by $i$, adding, and taking real and imaginary parts $$ u_x =v_y \\ u_y = -v_x $$ differentiating the first wrt $x$ and the second wrt $y$ and adding gives the harmonic property: $$ u_{xx}+u_{yy}=0 $$ Green's theorem has $$ \int_{\partial A} Pdx+Qdy = \int_A (Q_x-P_y)dS $$ setting $P=u_y,Q=-u_x$ $$ \int_{\partial A} u_y dx -u_xdy = -\int_A(u_{xx}+u_{yy})dS = 0 $$