Verify that each given function $u$ is harmonic (in the region where it is defined) and then find a harmonic conjugate of $u$.
(a) $u=y$
I was able to verify that $u$ is harmonic pretty easily. But I found that the harmonic conjugate of $u$ is $-x+a$. Thus, I get that $f(z)= y + i(-x+a)$. Using that $y = \frac{z-\bar{z}}{2i}$ and that $x=\frac{z+\bar{z}}{2}$, I get that $f(z) = -i(\bar{z}+a)$, which as I understand is not an analytic function. I would like to know if $f(z)$ is indeed analytic, and whether or not having a harmonic function $u$ along with its conjugate $v$ in $f(z)$ guarantees having an analytic function $f(z)$?
Suppose $D \subset \Bbb C$ is a domain and $u : D \to \Bbb R$ is a harmonic function.
If $D$ is simply connected, then $u$ admits a harmonic conjugate (globally). Explicitly, up to an overall constant it is (for any choice of base point $z_0 \in D$) $$v(z) := \int_{z_0}^z -u_y \,dx + u_x \,dy .$$ Computing the exterior derivative of the integrated $1$-form gives $$d(-u_y \,dx + u_x \,dy) = (u_{xx} + u_{yy}) \, dx \wedge dy = 0 ,$$ where the second equality just uses that $u$ is harmonic, so $-u_y \,dx + u_x \,dy$ is closed and hence (since $D$ is simply connected) exact, and thus the integral is well-defined. If $-u_y \,dx + u_x \,dy = df$, then $v(z) = \int_{z_0}^z df = f(z) - f(z_0)$.
Now, differentiating shows that the pair $(u, v)$ satisfies the Cauchy-Riemann equations---in fact the definition of $v$ is rigged precisely so that this happens---or equivalently that the function $f : D \to \Bbb C$, $f := u + i v$, is analytic on $D$.
On the other hand, if $D$ is not simply connected, $u$ may not admit a global harmonic conjugate, i.e., one on all of $D$, whereas the above argument shows that for any simply connected subset $A \subset D$, $u\vert_A$ does admit a harmonic conjugate $v : A \to \Bbb R$. To find a counterexample, inspecting the argument shows that we certainly need a $1$-form that is closed but not exact (any such $1$-form must have domain that is not simply connected). The standard example is $$\frac{-y \,dx + x \,dy}{x^2 + y^2}.$$ On any simply connected subset of $\Bbb C - \{ 0 \}$ this is the exterior derivative of some function, and the Cauchy-Riemann equations imply that (the negative of) its harmonic conjugate has exterior derivative $$\frac{x \,dx + y \,dy}{x^2 + y^2} .$$ But this expression defines a $1$-form on all of $\Bbb C - \{ 0 \}$, and this $1$-form is exact: An easy integration shows that it's just the exterior derivative of $$u(z) := \frac{1}{2} \log (x^2 + y^2) = \log |z| .$$ By construction, $u$ does not have a harmonic conjugate on all of $\Bbb C - \{ 0 \}$. Up to additive constants, the harmonic conjugates of the restrictions of $u$ to simply connected subsets are branches of the argument function.