I am working through a System and Signals book on my own time, and have acquired a copy of the answer key to check my work.
One of the problems involves proving $(\vert z_1 \vert – \vert z_2 \vert)^2 \leq \vert z_1 + z_2 \vert^2$. The strategy the answer key uses relies on the identity $\vert z_1 + z_2 \vert^2 = r_1^2 + r_2^2 + 2r_1r_2\cos(\theta_1 – \theta_2)$, (where $r_1$ and $\theta_1$ are the magnitude and angle of $z_1$) which it simply declares without any further elaboration.
How does one get this identity? Is this well-known?
What I've tried:
Knowing that I probably ought to stay in polar form, I start with… $$\vert z_1 + z_2 \vert^2 = \vert r_1e^{j\theta_1} + r_2e^{j\theta_2} \vert^2 = \vert r_1^2e^{j2\theta_1} + r_2^2e^{j2\theta_2} + 2r_1r_2e^{j(\theta_1 + \theta_2)} \vert$$
…which I can then divide out $e^{j(\theta_1 + \theta_2)}$ with…
$$
\begin{align}
\vert r_1^2e^{j2\theta_1} + r_2^2e^{j2\theta_2} + 2r_1r_2e^{j(\theta_1 + \theta_2)} \vert &=
\vert r_1^2e^{j(\theta_1 – \theta_2)} + r_2^2e^{-j(\theta_1 – \theta_2)} + 2r_1r_2\vert \cdot \vert e^{j(\theta_1 + \theta_2)}\vert \\
&= \vert r_1^2e^{j(\theta_1 – \theta_2)} + r_2^2e^{-j(\theta_1 – \theta_2)} + 2r_1r_2\vert
\end{align}
$$
This seems to be getting closer to a cosine being inside the equation, but at the same time not where I would expect.
Best Answer
My most immediate inclination is to use the inner product. Recall, if $z := a+ib, w := c+id \in \mathbb{C}$ then we may define $$ \newcommand{\ip}[1]{\left\langle #1 \right\rangle} \ip{z,w} := \overline{z} w = (a-ib)(c+id) = (ac+bd) + i(ad-bc) $$ We can define norms via inner products (in general), too: $$ \| z \| := \sqrt{\ip{z,z}} $$ (We call this the norm induced by the inner product. For complex numbers, this is the modulus, $|z|$.)
An identity that holds for inner products in general (which you can prove from the definitions of such) is $$ \|x+y\|^2 = \ip{x+y,x+y} = \|x\|^2 + \|y\|^2 + 2 \cdot \mathfrak{Re} \ip{x,y} $$ To prove this, expand out $\ip{x+y,x+y}$ using the sesqui-linearity of the inner product, and the conjugate-symmetry rule of $\ip{x,y} = \overline{\ip{y,x}}$. If you wish to prove it for the given inner product described at the start without appealing to these rules, you may note that $$ \overline{z+w} = \overline z + \overline w $$ and $$ |z|^2 = \overline z z $$ and $$ \ip{z+w,z+w} = \overline{(z+w)} (z+w) $$
Of course, then, the question is how to rectify this $\mathfrak{Re} \ip{x,y}$ term into something useful. For that, simply find $\ip{x,y}$, express the result in the polar form, and take the real part according to Euler's formula, $e^{i\theta} = \cos \theta + i \sin \theta$.