My problem is the following:
Let $u$ be a continuous real-valued function in the closure of the unit disk $\mathbb{D}$ that is harmonic in $\mathbb{D}$. Assume that the boundary values of $u$ are given by $ u(e^{it}) = 5- 4 \cos t. $ Furthermore, let $v$ be a harmonic conjugate of $u$ in $\mathbb{D}$ such that $v(0) = 1$. Find $u(1/2)$ and $v(1/2)$.
It's easy to find $u(1/2)$ using the Poisson integral formula: $u(z) = \frac{1}{2\pi} \int_0 ^{2\pi} \frac{1-|z|^2}{|e^{i\theta}-z|^2} u(e^{i\theta}) d\theta$ yields $u(1/2) = \frac{1}{2 \pi} \int_0^{2\pi} \frac{3/4}{5/4-\cos \theta} (5-4\cos \theta) d\theta = \frac{1}{2\pi} \int_0^{2\pi} 3 d\theta = 3.$
I get stuck trying to find the value for $v$. I know that $ 1 = v(0) = \frac{1}{2\pi} \int_0^{2\pi} v(e^{i\theta}) d\theta.$ Also, $ v(1/2) = \frac{1}{2\pi} \int_0^{2\pi} \frac{3/4}{5/4 - \cos \theta} v(e^{i\theta}) d\theta = \frac{1}{2\pi} \int_0^{2\pi} \frac{3v(e^{i\theta})}{u(e^{i\theta})} d\theta, $ and in general, the harmonic conjugate is given by the line integral $ v(z) = \mathcal{Im} \int_0^z f'(w) dw + C.$ I don't know how to proceed with this information to find the $v$-value.