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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.

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    $\begingroup$ Expand using $|z|^2 = \bar{z} z$. $\endgroup$
    – copper.hat
    Jun 6, 2023 at 4:18
  • $\begingroup$ @copper.hat This answer let me figure out how the book would have expected someone at my skill level to solve it. You should make it its own answer. $\endgroup$ Jun 8, 2023 at 22:52
  • $\begingroup$ Note that $|z|^2 = \overline{z} z$, this is easy to show by writing $z=x+iy$ and multiplying. $|z_1+z_2|^2 = (\overline{z_1+z_2}) (z_1+z_2) = |z_1|^2+|z_2|^2 + 2 \operatorname{re} z_1 \overline{z_2}$. Use $z_k = r_k e^{i \theta_k}$ to show that $\operatorname{re} z_1 \overline{z_2} = r_1r_2 \cos ( \theta_1-\theta_2)$. $\endgroup$
    – copper.hat
    Jun 8, 2023 at 23:15

3 Answers 3

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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, $e^{i\theta} = \cos \theta + i \sin \theta$.

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  • $\begingroup$ This is a very beautiful proof, but the textbook I'm using begins by covering what complex numbers even are. I didn't even know what "sesqui-linearity" was until looking it up. That being said it does work! $\endgroup$ Jun 8, 2023 at 22:58
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As an alternative approach, I would recommend you to apply the triangle inequality twice.

To begin with, notice that \begin{align*} |z_{1}| = |(z_{1} + z_{2}) - z_{2}| \leq |z_{1} + z_{2}| + |z_{2}| \end{align*} Similarly, one also concludes that \begin{align*} |z_{2}| = |(z_{1} + z_{2}) - z_{1}| \leq |z_{1} + z_{2}| + |z_{1}| \end{align*}

Gathering both results, it results that: \begin{align*} |z_{1} + z_{2}| \geq \max\{|z_{1}| - |z_{2}|, |z_{2}| - |z_{1}|\} = ||z_{1}| - |z_{2}|| \end{align*}

and we are done.

Hopefully this helps!

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We can also show this geometrically by plotting the complex numbers as vectors on the Argand plane. Let $O$ be the origin, $A$ be the point represented by $z_1$, $B$ be the point represented by $z_2$ and $C$ be the point represented by $z_1 + z_2$.

We can see that the length of $OC$ is $|z_1+z_2|$. By chasing angles, you can show that $\angle OBC$ is equal to $\pi - (\theta_1 - \theta_2)$. Applying the cosine rule on $\triangle OBC$ gives you $$ |z_1+z_2|^2 = r_1^2 + r_2^2 - 2 r_1 r_2 \cos(\pi - (\theta_1 - \theta_2)),$$ and the result follows after applying the identity $\cos (\pi - \alpha) = - \cos\alpha$.

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