Find the function $u_{j}(x,y)$ and $v_{j}(x,y) $ so that $w_{j}(x,y)+iv_{j}(x,y)$ and $u_{j}(x,y)+iw_{j}(x,y)$ My function  $w_{j}(x,y)=x+3y$ is harmonic
$w_{xx}+w_{yy}=0+0$
In order to find $v_{1}(x,y)$ I use Cauchy-Riemann
$v_{y}=w_{x}$ and $v_{x}=-w_{y}$ 
So for $v_{y}$ and $v_{x}$
$v_{1}(x,y)=\int1dy=x+A(X)$
$v_{1}(x,y)=\int-3dx=-3x+B(x)$
And from here I am stuck. My textbook doesn't explain harmonic conjugates very well, so any help will be most grateful!

Given the harmonic function
$w = x + 3y, \tag 0$
we have, by the Cauchy-Riemann equations,
$v_x = -w_y = -3, \tag 1$
and
$v_y = w_x = 1; \tag 2$
thus
$\nabla v = \begin{pmatrix} v_x \\ v_y \end{pmatrix} = \begin{pmatrix} -3 \\ 1 \end{pmatrix}; \tag 3$
it follows that, given any two points $(x_0, y_0), (x, y) \in \Bbb R^2$, and any differentiable path $\gamma(t)$ 'twixt $(x_0, y_0)$ and $(x, y)$,
$\gamma(t): [0, 1] \to \Bbb R^2; \; \gamma(0) = (x_0, y_0), \; \gamma(1) = (x, y), \tag 4$
that
$v(x, y) - v(x_0, y_0) = \displaystyle \int_0^1 \dfrac{dv(\gamma(s))}{ds} \; ds = \int_0^1 \nabla v(\gamma(s)) \cdot \gamma'(s) \; ds; \tag 5$
according to (3), the vector field $\nabla v$ will be constant along any such curve $\gamma(t)$; thus
$v(x, y) - v(x_0, y_0) = \displaystyle \int_0^1 \begin{pmatrix} -3 \\ 1 \end{pmatrix} \cdot \gamma'(s) \; ds; \tag 6$
if we take $\gamma(t)$ to be a line segment joining $(x_0, y_0)$ and $(x, y)$,
$\gamma(t) = t(x, y) + (1 - t)(x_0, y_0) = (tx + (1 - t)x_0, ty + (1 - t)y_0)$
$= (x_0 + t(x - x_0), y_0 + t(y - y_0)), \tag 7$
then
$\gamma'(t) = \begin{pmatrix} x - x_0 \\ y - y_0 \end{pmatrix}; \tag 8$
the integral (6) thus becomes
$v(x, y) - v(x_0, y_0) = \displaystyle \int_0^1 \begin{pmatrix} -3 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} x - x_0 \\ y - y_0 \end{pmatrix} \; ds = \displaystyle \int_0^1 (-3(x - x_0) + (y - y_0)) \; ds$
$= \displaystyle (-3(x - x_0) + (y - y_0)) \int_0^1 ds = -3(x - x_0) + (y - y_0) = -3x + y + (3x_0 - y_0); \tag 9$
the expression $-3(x - x_0) + (y - y_0)$ may brought outside the integral since it is constant with respect to the variable of integration $s$; thus
$v(x, y) = -3x + y + (3x_0 - y_0) + v(x_0 y_0); \tag {10}$
$v(x_0, y_0)$ may be chosen freely and we may set
$C = (3x_0 - y_0) + v(x_0, y_0), \tag{11}$
and it follows that the harmonic conjugate of $w(x, y) = x + 3y$ is of the form
$v(x, y) = -3x + y + C; \tag{12}$
it follows that
$w(x, y) + iv(x, y) = x + 3y + i(-3x + y + C) \tag{13}$
is holomorphic; the reader may easily check that it obeys the Cauchy-Riemann equations
Having found $v$ such that $w + iv$ is holomorphic, we observe that
$-v + iw = i(w + iv) \tag{14}$
is also holomorphic; thus taking
$u = -v \tag{15}$
we find that $u + iw$ is holomorphic as well.
