Intuitive "conversion" between the law of cosines in usual geometry and complex analysis My question is the converse of this one.
So, can we use the law of cosines $$c^2=a^2+b^2-2ab \cos \theta$$ to get the intuition of the complex law of cosines expressed as $$|z+w|^2=|z|^2+|w|^2+2 \Re(z \bar w)$$
My problem is to find the intuitive "conversion" between the two that helps me to remember it. Note that there are other version of the formula for the complex case that may be better.
 A: Geometry First!
Allow me to impress you with my MSPaint-equivalent skills. Here, we're just using standard vector geometry to draw things.



*

*Note that $\cos(\theta) = -\cos(\pi - \theta)$ by a basic identity.

*Recall that the dot product of two vectors $\mathbf{a}, \mathbf{b}$ can be interpreted geometrically by $\mathbf{a} \cdot \mathbf{b} = \lVert \mathbf{a} \rVert \lVert \mathbf{b} \rVert \cos \theta$, where $\theta$ is the angle between the two vectors. 
Now, let's think about that interesting term involving the real part of $z \bar{w}$. For calculations we'll write $z = x_z + iy_z$, while for geometry's sake we can think of $z$ as the real vector $\begin{bmatrix}x_z \\ y_z\end{bmatrix}$ (similarly for $w$). 
We have
$$\operatorname{Re}(z\bar{w}) = \operatorname{Re}\big((x_z + iy_z)(x_w - iy_w)\big) = x_zx_w + y_zy_w,$$ which is exactly the dot product $\begin{bmatrix}x_z\\ y_z\end{bmatrix} \cdot \begin{bmatrix}x_w\\ y_w\end{bmatrix}$, and so must be $\lvert z \rvert \lvert w \rvert \cos \theta$.
All together,
\begin{align*}
\lvert z + w \rvert 
&= \lvert z \rvert^2 +  \lvert w \rvert^2 - 2\lvert z \rvert \lvert w \rvert \cos (\pi - \theta) \tag{by picture} \\
&= \lvert z \rvert^2 +  \lvert w \rvert^2 + 2\lvert z \rvert \lvert w \rvert \cos \theta \tag{cosine identity} \\
&= \lvert z \rvert^2 +  \lvert w \rvert^2 + 2\operatorname{Re}(z\bar{w}). \tag{remarks on $\operatorname{Re}(z\bar{w}$)} \\
\end{align*}
And some algebra, for good measure
Another approach to take is based on combining standard algebraic identities in a rather "inspired" way. We'll need to know that:


*

*For complex $z$, we have $\lvert z \rvert^2 = z\bar{z}$,

*Conjugation is nice in several ways. Additively, $\overline{z + w} = \bar z + \bar w$, multiplicatively, $\overline{zw} = \bar z \bar w$, and self-inverse in that $\overline{\bar{z}} = z$.

*$z + \bar{z} = 2\operatorname{Re}(z)$.
Now we'll just play with $\lvert z + w \rvert$.
\begin{align*}
\lvert z + w \rvert 
&= (z + w)(\overline{z + w}) \\
&= (z + w)(\bar z + \bar w) \\
&= z\bar z + z \bar w + w \bar z + w \bar w \\
&= \lvert z \rvert^2 + \lvert w \rvert^2 + z \bar w + \bar z w \\
&= \lvert z \rvert^2 + \lvert w \rvert^2 + z \bar w + \overline{z \bar w} \\
&= \lvert z \rvert^2 + \lvert w \rvert^2 + 2\operatorname{Re}(z \bar w) 
\end{align*}
I think the geometry is really better for intuition, but if you're comfortable with how nice conjugation is, and its role in defining the inner product, it's not too hard to fumble from $\lvert z + w \rvert$ to what we want (it's what I did!).
