I have a very silly question to ask! I have

$|z_{1} + z_{2}|^2 = |z_{1}|^2+|z_{2}|^2+2|z_{1}||z_{2}|\cos{\theta}$,

where $z_{1}$ and $z_{2}$ are complex numbers.

For the life of me I cannot reproduce the $\cos{\theta}$ term, to me this looks like the geometric definition of the dot-product but I am unsure as to whether you can treat a complex number in the same way as a "traditional" vector.

I also tried to use Euler's identity to see if the $\cos{\theta}$ term could be produced via its exponential form - no luck!

Any thoughts are appreciated!

  • $\begingroup$ what's $\theta$ ? $\endgroup$ – mercio Jan 11 '16 at 2:02
  • $\begingroup$ Sorry I should have said, the phase difference between $z_{1}$ and $z_{2}$. Something like $\theta_{1}-{\theta_{2}}$? $\endgroup$ – QuantumPenguin Jan 11 '16 at 2:06

Use conjugates: $$|z_1+z_2|^2 =(z_1+z_2)(\overline{z_1}+\overline{z_2}) =z_1\overline{z_1}+z_2\overline{z_2}+z_1\overline{z_2}+\overline{z_1}z_2\ ,$$ that is, $$|z_1+z_2|^2 =|z_1|^2+|z_2|^2+2\Re(z_1\overline{z_2})\ .$$ Now if $z_1=re^{i\alpha}$ and $z_2=se^{i\beta}$ then $$\Re(z_1\overline{z_2})=\Re\{rse^{i(\alpha-\beta)}\}=rs\cos(\alpha-\beta)=|z_1||z_2|\cos\theta\ .$$

  • $\begingroup$ Ah, I see! Thanks! $\endgroup$ – QuantumPenguin Jan 11 '16 at 2:17
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    $\begingroup$ Hint: for something like this it is nearly always worth trying conjugates, because conjugates add up "nicely" while the modulus doesn't. $\endgroup$ – David Jan 11 '16 at 2:22
  • $\begingroup$ I will next time, I completely forgot that conjugates were applicable! $\endgroup$ – QuantumPenguin Jan 11 '16 at 2:53

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