number theory - $sum_{n=1}^infty chi(n)phi(n)n^{-s} = frac{L(chi,s-1)}{L(chi,s)}$

Let $\chi$ be a Dirichlet character mod 4. Show $\sum_{n=1}^\infty \chi(n)\phi(n)n^{-s} = \frac{L(\chi,s-1)}{L(\chi,s)}$ and $\sum_{n=1}^\infty \chi(n)d(n)n^{-s}=L(\chi,s)^2$. ($\phi$ is the Euler totient function and $d(n)$ is the number of divisors of $n$.)

First, is this true just for characters mod 4 and not true in general?
I'm not sure what specific properties about characters mod 4 I should use besides that $\chi(n)=0$ for $n$ even.

I took the log of both sides and tried to use the following:
$$L(\chi,s)=\prod_{p \text{ prime}}\frac{1}{1-\frac{\chi(p)}{p^s}}$$

$$\log L(\chi,s)=\sum_{p \text{ prime}}\sum_{n=1}^\infty \frac{\chi(p)^n}{np^{ns}}$$

$\chi$ and $\phi$ are multiplicative, so we can express $$\sum_{n=1}^\infty \chi(n)\phi(n)n^{-s}$$ as the Euler product $$\prod_p(1+\frac{\chi(p)\phi(p)}{p^s}+\frac{\chi(p^2)\phi(p^2)}{p^{2s}}+\cdots).$$

Manipulating things are not quite working. Any help would be appreciated.