Let $(M, g)$ be a Kähler manifold and $\rm{Ric}$ be its Ricci tensor defined on $T_\mathbb{C}M$ as the $\mathbb C$-bilinear extension of the usual Ricci tensor on $TM$. I want to prove that in any local frame $(Z_1,\ldots,Z_n, \bar{Z_1},\ldots,\bar{Z_n})$ of $T_\mathbb{C}M$ we have that $$ g^{\ell, \bar{\nu}} \mathrm{Ric}_{\mu, \bar{\nu}} =R^\ell_{ \alpha, \bar{\alpha}, \mu} .$$ I've seen the LHS of the desired inequality often written as $R^\ell_\mu$. So far, I've tried the following: using that $\mathrm{Ric}_{\mu, \bar{\nu}} = g^{\alpha, \bar{\beta}} \mathrm{Rm}_{\alpha, \bar{\beta}, \mu, \bar{\nu}}$ and $\mathrm{Rm}_{\alpha, \bar{\beta}, j, \bar{k}}=g_{\mu, \bar{\beta}}R^\mu_{j, \bar{k}, \alpha}$, we arrive at $$ g^{\ell, \bar{\nu}} \mathrm{Ric}_{\mu, \bar{\nu}} =g^{\ell, \bar{\nu}} g^{\alpha, \bar{\beta}} \mathrm{Rm}_{\alpha, \bar{\beta}, \mu, \bar{\nu}}= g^{\alpha, \bar{\beta}} \Big(g^{\ell, \bar{\nu}} \mathrm{Rm}_{\alpha, \bar{\beta}, \mu, \bar{\nu}}\Big) =g^{\alpha, \bar{\beta}} R^\ell_{ \alpha, \bar{\beta}, \mu }. $$ However, I can't see how to continue from here.
I must add that I'm using standard conventions here, namely: $ R(X,Y)Z=\nabla _{X}\nabla _{Y}Z-\nabla _{Y}\nabla _{X}Z-\nabla _{[X,Y]}Z $ where $X, Y, Z$ are complex vector fields, $\nabla$ is the complex $\mathbb C$-bilinear extension of the Levi-Civita connection of $(M, g)$ and $\mathrm{Rm}(X, Y, Z, W)=g(R(X, Y)Z, W)$. Any help is appreciated, regards.