# |z_1|/|z_2|=|z_1/z_2| proof (algebraic)

In this post I’ll be proving to you that:

$\frac { \left| { z }_{ 1 } \right| }{ \left| { z }_{ 2 } \right| } =\left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right|$

Firstly, I’ll say that:

${ z }_{ 1 }=x+iy,\quad \therefore \quad \left| { z }_{ 1 } \right| =\sqrt { { x }^{ 2 }+{ y }^{ 2 } }$

And also that:

${ z }_{ 2 }=p+iq,\quad \therefore \quad \left| { z }_{ 2 } \right| =\sqrt { { p }^{ 2 }+{ q }^{ 2 } }$

If this is the case, then:

$\frac { { z }_{ 1 } }{ { z }_{ 2 } } =\frac { x+iy }{ p+iq } \\ \\ =\frac { \left( x+iy \right) }{ \left( p+iq \right) } \cdot \frac { \left( p-iq \right) }{ \left( p-iq \right) } \\ \\ =\frac { px-iqx+ipy-{ i }^{ 2 }qy }{ { p }^{ 2 }-ipq+ipq-{ i }^{ 2 }{ q }^{ 2 } } \\ \\ =\frac { \left( px+qy \right) +i\left( py-qx \right) }{ { p }^{ 2 }+{ q }^{ 2 } } \\ \\ =\left( \frac { px+qy }{ { p }^{ 2 }+{ q }^{ 2 } } \right) +i\left( \frac { py-qx }{ { p }^{ 2 }+{ q }^{ 2 } } \right)$

And as this is in the form:

$z=a+ib$

I would have to conclude that:

$RHS=\left| \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right| \\ \\ =\sqrt { { \left( \frac { px+qy }{ { p }^{ 2 }+{ q }^{ 2 } } \right) }^{ 2 }+{ \left( \frac { py-qx }{ { p }^{ 2 }+{ q }^{ 2 } } \right) }^{ 2 } } \\ \\ =\sqrt { \frac { { \left( px+qy \right) }^{ 2 } }{ { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }^{ 2 } } +\frac { { \left( py-qx \right) }^{ 2 } }{ { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }^{ 2 } } } \\ \\ =\sqrt { \frac { { \left( px+qy \right) }^{ 2 }+{ \left( py-qx \right) }^{ 2 } }{ \left( { p }^{ 2 }+{ q }^{ 2 } \right) ^{ 2 } } } \\ \\ =\sqrt { \frac { { p }^{ 2 }{ x }^{ 2 }+2pqxy+{ q }^{ 2 }{ y }^{ 2 }+{ p }^{ 2 }{ y }^{ 2 }-2pqxy+{ q }^{ 2 }{ x }^{ 2 } }{ { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }^{ 2 } } } \\ \\ =\sqrt { \frac { { p }^{ 2 }{ x }^{ 2 }+{ q }^{ 2 }{ x }^{ 2 }+{ p }^{ 2 }{ y }^{ 2 }+{ q }^{ 2 }{ y }^{ 2 } }{ { \left( { p }^{ 2 }+{ q }^{ 2 } \right) }^{ 2 } } } \\ \\ =\sqrt { \frac { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \left( { p }^{ 2 }+{ q }^{ 2 } \right) }{ \left( { p }^{ 2 }+{ q }^{ 2 } \right) \left( { p }^{ 2 }+{ q }^{ 2 } \right) } } \\ \\ =\sqrt { \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { p }^{ 2 }+{ q }^{ 2 } } } \\ \\ =\frac { \sqrt { { x }^{ 2 }+{ y }^{ 2 } } }{ \sqrt { { p }^{ 2 }+{ q }^{ 2 } } } \\ \\ =\frac { \left| { z }_{ 1 } \right| }{ \left| { z }_{ 2 } \right| } =LHS$

Hence, I have my proof.

# How to prove that |z_1|*|z_2|=|z_1*z_2|, Complex Numbers

In this post I’ll be showing you how to prove that:

$\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| =\left| { z }_{ 1 }{ z }_{ 2 } \right|$

Firstly, let’s say that:

${ z }_{ 1 }=x+iy$

${ z }_{ 2 }=p+iq$

If this is the case, then according to the rules of complex numbers:

$\left| { z }_{ 1 } \right| =\sqrt { { x }^{ 2 }+{ y }^{ 2 } }$

$\left| { z }_{ 2 } \right| =\sqrt { { p }^{ 2 }+{ q }^{ 2 } }$

Secondly, let’s determine what ${ z }_{ 1 }{ z }_{ 2 }$ is…

${ z }_{ 1 }{ z }_{ 2 }\\ \\ =\left( x+iy \right) \left( p+iq \right) \\ \\ =px+iqx+ipy+{ i }^{ 2 }qy\\ \\ =px+iqx+ipy-qy\\ \\ =\left( px-qy \right) +i\left( qx+py \right)$

As you can see, we get the result above – which is another complex number.

This means that:

$RHS\\ \\ =\left| { z }_{ 1 }{ z }_{ 2 } \right| \\ \\ =\sqrt { { \left( px-qy \right) }^{ 2 }+{ \left( qx+py \right) }^{ 2 } } \\ \\ =\sqrt { \left( px-qy \right) \left( px-qy \right) +\left( qx+py \right) \left( qx+py \right) } \\ \\ =\sqrt { { p }^{ 2 }{ x }^{ 2 }-2pqxy+{ q }^{ 2 }{ y }^{ 2 }+{ q }^{ 2 }{ x }^{ 2 }+2pqxy+{ p }^{ 2 }{ y }^{ 2 } } \\ \\ =\sqrt { { p }^{ 2 }{ x }^{ 2 }+{ q }^{ 2 }{ y }^{ 2 }+{ q }^{ 2 }{ x }^{ 2 }+{ p }^{ 2 }{ y }^{ 2 } } \\ \\ =\sqrt { { p }^{ 2 }{ x }^{ 2 }+{ q }^{ 2 }{ x }^{ 2 }+{ p }^{ 2 }{ y }^{ 2 }+{ q }^{ 2 }{ y }^{ 2 } } \\ \\ =\sqrt { \left( { x }^{ 2 }+{ y }^{ 2 } \right) \left( { p }^{ 2 }+{ q }^{ 2 } \right) } \\ \\ =\sqrt { { x }^{ 2 }+{ y }^{ 2 } } \cdot \sqrt { { p }^{ 2 }+{ q }^{ 2 } } \\ \\ =\left| { z }_{ 1 } \right| { \left| { z }_{ 2 } \right| }\\ \\ =LHS$

Therefore we’ve proven that:

$\left| { z }_{ 1 } \right| \left| { z }_{ 2 } \right| =\left| { z }_{ 1 }{ z }_{ 2 } \right|$

You can watch a video related to this proof below…