# How to derive the identity $\vert z_1 + z_2 \vert^2 = r_1^2 + r_2^2 + 2r_1r_2\cos(\theta_1 - \theta_2)$

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.

• Expand using $|z|^2 = \bar{z} z$. Jun 6, 2023 at 4:18
• @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. Jun 8, 2023 at 22:52
• 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 $\operatorname{re} z_1 \overline{z_2} = r_1r_2 \cos ( \theta_1-\theta_2)$. Jun 8, 2023 at 23:15

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

• 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! Jun 8, 2023 at 22:58

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!

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 $$\cos (\pi - \alpha) = - \cos\alpha$$.