# Proving $|z_1z_2|=|z_1||z_2|$ using exponential form

Problem:

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 $$|z_1z_2|=|r_1e^{i\theta_1}\times r_2e^{i\theta_2}|= |r_1 r_2e^{i(\theta_1+\theta_2)}|$$ I cannot proceed further. Any help would be appreciated.

Hint:

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

• For all real valued $\theta$. Commented Aug 9, 2015 at 20:01
• Yes!!! I edit :) Commented Aug 9, 2015 at 20:03

$$|re^{i\theta}|=|r|$$

So

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

$$|z_{2}|=|r_{2}|$$

$$|z_{1}z_{2}|=|r_{1}r_{2}|$$

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

$$|z_{1}z_{2}|=|z_{1}||z_{2}|$$

• 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 $|z|=|r|$. Commented Aug 9, 2015 at 19:58
• Why does the first line follow? Commented Jan 18, 2023 at 20:49
• @Alper I assume that $\theta$ and $r$ are real numbers and so $re^{i\theta}=r\cos(\theta)+ir\sin(\theta)$. $|re^{i\theta}|=\sqrt{(r\cos(\theta))^{2}+(r\sin(\theta))^{2}}=|r|$. Commented Jan 19, 2023 at 21:05
• For a complex number $z=x+iy$, with $x$ and $y$ real we have $|z|=\sqrt{x^{2}+y^{2}}$ Commented Jan 19, 2023 at 21:06

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