# How to show $\text{Real}(z\sinh(-R + it)) = - \text{Real}(z) \cos(t) \text{sinh}(R) - \text{Imag}(z) \sin(t) \cosh(R)$?

I came across the following expression where $z$ is complex and $R$ and $t$ are real numbers: $\text{Real}(z\sinh(-R + it)) = - \text{Real}(z) \cos(t) \text{sinh}(R) - \text{Imag}(z) \sin(t) \cosh(R)$

I have never seen an identity like this before, how was it derived?

\begin{align} \sinh(-R+it)&=\frac{e^{-R+it}-e^{R-it}}2\\\\ &= \frac{e^{-R}(\cos(t)+i\sin(t))-e^R(\cos(t)-i\sin(t))}{2}\\\\ &=-\left(\frac{e^R-e^{-R}}2\right)\cos(t)+i\left(\frac{e^R+e^{-R}}2\right)\sin(t)\\\\ &=-\sinh(R)\cos(t)+i\cosh(R)\sin(t) \end{align}
Then, we note that $\text{Re}(iz)=-\text{Im}(z)$.
\begin{align} \text{Re}(z\sinh(-R+it))&=\text{Re}\left(-z\sinh(R)\cos(t)+iz\cosh(R)\sin(t)\right)\\\\ &=-\sinh(R)\cos(t)\text{Re}(z)-\cosh(R)\sin(t)\text{Im}(z)\end{align}