Prove $$|z_1z_2|=|z_1||z_2|$$ where $z_1,z_2$ are Complex Numbers.

I tried to solve this using the exponential form of a Complex Number.

Assuming $z_1=r_1e^{i\theta_1}$ and $z_2=r_2e^{i\theta_2},$ I got $$|z_1z_2|=|r_1e^{i\theta_1}\times r_2e^{i\theta_2}|= |r_1 r_2e^{i(\theta_1+\theta_2)}|$$ Unfortunately I cannot think of how to proceed further. Any help would be greatly appreciated! Many thanks in anticipation!




$$|z_{1}|=|r_{1}| $$



$r_{1},r_{2}$ are real numbers and so $|r_{1}r_{2}|=|r_{1}||r_{2}|=|z_{1}||z_{2}|$


  • $\begingroup$ Sir, how is $|re^{i\theta}|=|r|?$ $\endgroup$ – Ishan Aug 9 '15 at 19:55
  • $\begingroup$ Let $z=r e^{i\theta}$, the complex conjugate of $z$ ,$z^{*}=re^{-i\theta}$, $|z|^{2}=zz^{*}=r^{2}e^{i\theta-i\theta}=r^{2}e^{0}=r^{2}$, then taking the positive square root of both sides since we are discussing a magnitude of this complex number $|z|=|r|$. $\endgroup$ – Tucker Aug 9 '15 at 19:58


$\forall \theta \in \mathbb{R}$ we have. $$ |e^{i\theta}|=|\cos \theta + i \sin \theta|=\cos^2 \theta +\sin^2 \theta=1 $$

  • 1
    $\begingroup$ For all real valued $\theta$. $\endgroup$ – Tucker Aug 9 '15 at 20:01
  • $\begingroup$ Yes!!! I edit :) $\endgroup$ – Emilio Novati Aug 9 '15 at 20:03

You are almost there. $|r_1r_2e^{i(\theta_1+\theta_2)}|=r_1r_2=|z_1||z_2|$


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