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By using the polar form of the complex number prove that, $|z_1 z_2| = |z_1| |z_2|$ and $\left|\frac{z_1}{z_2}\right| = \frac{|z_1|}{|z_2|}$

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closed as off-topic by Matthew Conroy, Avitus, Michael Albanese, Shailesh, Leucippus Jul 14 '16 at 2:41

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Hint:

remember that $|e^{i\theta}|=1$ and use the polar decomposition: $$ z_1=|z_1|e^{i\theta_1} \qquad z_2=|z_2|e^{i\theta_2} $$

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Using polar form, let $$z_1 = r_1(\cos{\theta_1} + i \sin{\theta_1})$$ $$z_2 = r_2(\cos{\theta_2} + i \sin{\theta_2})$$ Then, we have: $$|z_1 \cdot z_2| = |r_1r_2(\cos{\theta_1}\cos{\theta_2} - \sin{\theta_1}\sin{\theta_2} + i(\sin{\theta_1}\cos{\theta_2} + \cos{\theta_1}\sin{\theta_2}))| \\ = |r_1 r_2 (\cos(\theta_1 + \theta_2)+ i\sin(\theta_1 + \theta_2))| = r_1 r_2 $$ where I used the trignometric addition identities. Therefore, $|z_1 \cdot z_2| = |z_1| \cdot |z_2|$.

Similarly, $$ \left|\frac{z_1}{z_2}\right| = \left| \frac{r_1(\cos{\theta_1} + i \sin{\theta_1})}{r_2(\cos{\theta_2} + i \sin{\theta_2})} \right| = \left| \frac{r_1(\cos{\theta_1} + i \sin{\theta_1})(\cos{\theta_2} - i \sin{\theta_2})}{r_2(\cos^2{\theta_2} + \sin^2{\theta_2})} \right| = \left| \frac{r_1}{r_2}(\cos{\theta_1}\cos{\theta_2} + \sin{\theta_1} \sin{\theta_2} + i(\sin{\theta_1}\cos{\theta_2} - \cos{\theta_1} \sin{\theta_2})) \right| = \left| \frac{r_1}{r_2}\left(\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)\right) \right| = \frac{r_1}{r_2} = \frac{|z_1|}{|z_2|}$$

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The absolute value of a number is its distance from $0$. Thus, all absolute values are either positive or 0. That is, if we have a number $x$, then the absolute value of x which can be written as $|x|$ is equal to:

  1. $x$ if $x$ is positive
  2. $-x$ if $x$ is negative
  3. $0$ if $x$ is $0$

Also, for any real number, $x^2$ is positive or $0$ (if $x=0$). Therefore, $\sqrt{x^2}$ is:

  1. $x$ if $x$ is positive
  2. $-x$ if $x$ is negative
  3. $0$ if $x$ is $0$

If the $-x$ part is a bit confusing, consider take for example:

$$\sqrt{(-8)^2}=-(-8)=8$$

From the discussion above we can conclude that $|x|=\sqrt{x^2}$.

Using this fact we are going to prove to arguments in this post.

If $a$ and $b$ are two real numbers:

(1) the absolute value of their product is equal to the product of their absolute values or

$$|ab|=|a||b|$$

and

(2) the absolute value of their quotient is equal to the quotient of their absolute values or

$$\left|\frac{a}{b}\right|=\frac{|a|}{|b|}$$

Proof of (1)

From above, we know that $|x|=\sqrt{x^2}$, so

$$|ab|=\sqrt{(ab)^2}=\sqrt{a^2b^2}=\sqrt{a^2}\sqrt{b^2}$$

But from the definition above:

$$\sqrt{a^2}\sqrt{b^2}=|a||b|$$

Therefore, $|ab|=|a||b|$.

Proof of (2)

$$\left|\frac{a}{b}\right|=\sqrt{\left(\frac{a}{b}\right)^2}=\frac{\sqrt{a^2}}{\sqrt{b^2}}$$

Since $\sqrt{x^2}=|x|$:

$$\frac{\sqrt{a^2}}{\sqrt{b^2}}=\frac{|a|}{|b|}$$

Therefore,

$$\left|\frac{a}{b}\right|=\frac{|a|}{|b|}$$

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