How to establish the following result: $a\sin A=b\sin B\operatorname{cis}(∓C)+c\sin C\operatorname{cis}(±B)$ As the title suggests, I am trying to understand does one obtain the following relation to obtain a third vertex, say $a$, given that the angles $A,B,C$ of a triangle, with two of the vertices $b,c$ as complex numbers are known. Here, we assume that $A$ is not a multiple of $\pi$ (else there is no unique answer).  
Then, the answer is found to be $$a\sin A=b\sin B\operatorname{cis}(∓C)+c\sin C\operatorname{cis}(±B),$$
where $\operatorname{cis}\theta = e^{i\theta}$ with the upper signs if $a,b,c$ run anticlockwise, and the lower signs if clockwise.
This is used by the author in the proof of Napoleon's Theorem as seen here: https://www.cantab.net/users/michael.behrend/problems/morley_tri/pages/proof.html. 
Can someone kindly show me how can I prove the above relation as I am trying to understand every single step.   
Otherwise, can someone tell me the name of the theorem and I can try to research for its proof. Thank you once again.
 A: The idea is that you can obtain vector $\overrightarrow{ba}$ (that is, $a-b$) from $\overrightarrow{bc}$ (that is, $c-b$) via


*

*Rotation through angle $\pm B$ (depending upon clockwise-ness)

*Scaling by $|a-b|/|c-b|$


Now, $(1)$ is accomplished with multiplication by $\operatorname{cis}(\pm B)$. By the Law of Sines, $(2)$ is accomplished with multiplication by $\sin C/\sin A$. Thus,
$$\begin{align}
a - b &= ( c - b )\operatorname{cis}(\pm B) \;\frac{\sin C}{\sin A}\\[4pt]
\to\qquad ( a - b )\sin A &= ( c - b ) \operatorname{cis}(\pm B) \sin C \\[4pt]
\to\quad a \sin A &= b \left(\; \sin A - \operatorname{cis}(\pm B) \sin C \;\right) + c\operatorname{cis}(\pm B) \sin C
\end{align}$$ 
We need only show that that the multiplier on $b$ has the desired form.
$$\begin{align}
\sin A - \operatorname{cis}(\pm B)\sin C &= \sin A - (\cos B \pm i \sin B)\sin C \\
&= (\sin A - \cos B \sin C ) \mp i \sin B \sin C \\
&= (\sin(180^\circ - B-C) - \cos B \sin C ) \mp i \sin B \sin C \\
&= (\sin(B+C) - \cos B \sin C ) \mp i \sin B \sin C \\
&= \sin B \cos C \mp i \sin B \sin C \\
&= \sin B (\cos C \mp i \sin C ) \\
&= \sin B \operatorname{cis}(\mp C)
\end{align}$$
