Showing that the curves defined in the $xy$ plane by $u(x,y)=1$ and $v(x,y)=1$ cross at right angles at the origin. Suppose $f$ is an entire function with $f(0)=1+i$. Let $u(x,y)=Re(f(x+iy))$ and $v(x,y)=Im(f(x+iy))$.
A) Show that the function $u$ is a harmonic function of $x$ and $y$. 
B) Show that the curves defined in the $xy$ plane by $u(x,y)=1$ and $v(x,y)=1$ cross at right angles at the origin.
My Approach for A
I know I have to use the Cauchy Reumann Equations here but how? Am I suppose to show that for some $f: \mathbb{C} \to \mathbb{C}$, $u_{xx}+u_{yy}=0$?
My Approach for B
Is there a dot product involved here? 
Any help is appreciated. Thanks.

For $A$, use the Cauchy Riemann Eq : $u_x = v_y$ , $u_y = - v_x$ So
$$ u_{xx} = v_{yx} = (v_{x})_y = - ( u _y )_y \implies u_{xx} + u_{yy} = 0$$
For $B$, the curves with $u=1$ mean $f(x,y) = 1 + i v(x,y)$, curves with $v=1$ mean $f(x,y) = u(x,y) + i$. What can you say about these curves if $f(0,0) =1 + i$
