If $\mu$ is a signed measure with Jordan decomposition $\mu=\mu^+-\mu^-$, then the total variation measure of $\mu$ is equal to $\mu^++\mu^-$. My question is, is it similarly possible to express the total variation measure of a complex measure $\mu$ in terms of its real and imaginary parts $\mu_r$ and $\mu_i$?
Maybe $|\mu_r|^2 + |\mu_i|^2$ or the square root of that or something?
The short answer is no. To begin with, the total variation of a complex measure is defined as $$|\mu|(E):=\sup\left\{\sum_{j\in J}|\mu(A_j)|:\{A_j\}_{j\in J}\text{ is a disjoint measurable partition of }E\right\}.$$
Now, here's a counterexample to your question. In $\mathbb{R}$ with Borel sigma algebra, consider the measure $\mu=\delta_1-(1+i)\delta_0$. Thus, according to your notation, $\mu_r=\delta_1-\delta_0$ and $\mu_i=-\delta_0$. Then $|\mu_r(\{0,1\})|=2$ and $|\mu_i(\{0,1\})|=1$ but $$|\mu|(\{0,1\})=|\mu(\{0\})|+|\mu(\{1\})|=\sqrt{2}+1.$$