# 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…

# 2 ways to derive Pythagoras’ equation from scratch

The other day I discovered one more way to derive Pythagoras’ equation from scratch, completely by accident. I was deriving Pythagoras’ equation using the usual method, whilst navigating  a diagram similar to the one below, but without (B-A) measurements…

*Note (regarding diagram above): x+y = 90 degrees

The usual method goes like this…

The area of the largest square is:

${ \left( A+B \right) }^{ 2 }$

It is also:

$4\cdot \frac { 1 }{ 2 } AB+{ C }^{ 2 }$

Which means that:

${ \left( A+B \right) }^{ 2 }=4\cdot \frac { 1 }{ 2 } AB+{ C }^{ 2 }\\ \\ { A }^{ 2 }+2AB+{ B }^{ 2 }=2AB+{ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }$

Now, when I added the lengths (B-A) to my diagram, which are included in the diagram above, I discovered a new way to derive Pythagoras’ equation…

I did this by focusing on the area C^2. It turns out that:

$4\cdot \frac { 1 }{ 2 } AB+{ \left( B-A \right) }^{ 2 }={ C }^{ 2 }$

And since:

${ \left( B-A \right) }^{ 2 }\\ \\ ={ \left( A+B \right) }^{ 2 }-4AB\\ \\ ={ A }^{ 2 }+2AB+{ B }^{ 2 }-4AB\\ \\ ={ B }^{ 2 }-2AB+{ A }^{ 2 }$

I was able to say that:

$4\cdot \frac { 1 }{ 2 } AB+\left\{ { B }^{ 2 }-2AB+{ A }^{ 2 } \right\} ={ C }^{ 2 }\\ \\ 2AB+{ B }^{ 2 }-2AB+{ A }^{ 2 }={ C }^{ 2 }\\ \\ \therefore \quad { A }^{ 2 }+{ B }^{ 2 }={ C }^{ 2 }$

Obviously, I was quite pleased. Have you discovered other ways in which to derive Pythagoras’ equation??

Related:

Video on how to come up with Pythagoras’s equation…

How To Come Up With Pythagoras’s Equation