# Proving that arg(z_1/z_2)=arg(z_1)-arg(z_2)

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

$arg\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) =arg\left( { z }_{ 1 } \right) -arg\left( { z }_{ 2 } \right)$

Now firstly I will have to say that:

${ z }_{ 1 }={ r }_{ 1 }\left( \cos { { \theta }_{ 1 }+i\sin { { \theta }_{ 1 } } } \right) \\ \\ \therefore \quad arg\left( { z }_{ 1 } \right) ={ \theta }_{ 1 }$

And also that:

${ z }_{ 2 }={ r }_{ 2 }\left( \cos { { \theta }_{ 2 }+i\sin { { \theta }_{ 2 } } } \right) \\ \\ \therefore \quad arg\left( { z }_{ 2 } \right) ={ \theta }_{ 2 }$

If this is the case, then…

Since this is in the form:

$z=r\left( \cos { \theta +i\sin { \theta } } \right)$

I would have to conclude that:

$arg\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) ={ \theta }_{ 1 }-{ \theta }_{ 2 }=arg\left( { z }_{ 1 } \right) -arg\left( { z }_{ 2 } \right)$

Hence I’ve proven that:

$arg\left( \frac { { z }_{ 1 } }{ { z }_{ 2 } } \right) =arg\left( { z }_{ 1 } \right) -arg\left( { z }_{ 2 } \right)$

# How to derive Euler’s Identity using the Maclaurin Series

In this post I’ll be showing you how to derive Euler’s identity using the Maclaurin Series. It turns out that the Maclaurin series looks like this:

And expanded, it looks like this:

[*A larger version of this image can be found here.]

Now, since we want to derive Euler’s identity, we first have to find out what the formula for e^x looks like. In order to get this formula we must use the table below:

Derivatives of e^x When x=0
${ f }^{ \left( 0 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 0 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 1 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 1 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 2 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 2 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 3 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 3 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 4 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 4 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 5 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 5 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 6 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 6 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 7 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 7 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 8 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 8 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 9 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 9 \right) }\left( 0 \right) ={ e }^{ 0 }=1$
${ f }^{ \left( 10 \right) }\left( x \right) ={ e }^{ x }$ ${ f }^{ \left( 10 \right) }\left( 0 \right) ={ e }^{ 0 }=1$

Ok. So we’ve got a useful table just above. Let’s write out the function of e^x in its Maclaurin Series form:

$f\left( x \right) ={ e }^{ x }\\ \\ =\frac { 1 }{ 0! } { f }^{ \left( 0 \right) }\left( 0 \right) { x }^{ 0 }+\frac { 1 }{ 1! } { f }^{ \left( 1 \right) }\left( 0 \right) { x }^{ 1 }+\frac { 1 }{ 2! } { f }^{ \left( 2 \right) }\left( 0 \right) { x }^{ 2 }\\ \\ +\frac { 1 }{ 3! } { f }^{ \left( 3 \right) }\left( 0 \right) { x }^{ 3 }+\frac { 1 }{ 4! } { f }^{ \left( 4 \right) }\left( 0 \right) { x }^{ 4 }+\frac { 1 }{ 5! } { f }^{ \left( 5 \right) }\left( 0 \right) { x }^{ 5 }\\ \\ +\frac { 1 }{ 6! } { f }^{ \left( 6 \right) }\left( 0 \right) { x }^{ 6 }+\frac { 1 }{ 7! } f^{ \left( 7 \right) }\left( 0 \right) { x }^{ 7 }+\frac { 1 }{ 8! } { f }^{ \left( 8 \right) }\left( 0 \right) { x }^{ 8 }\\ \\ +\frac { 1 }{ 9! } { f }^{ \left( 9 \right) }\left( 0 \right) { x }^{ 9 }+\frac { 1 }{ 10! } { f }^{ \left( 10 \right) }\left( 0 \right) { x }^{ 10 }+...$

Now, let’s replace

with the values from the table. If we do this, the formula for e^x will become:

Alright, so far, so good. We are certainly on the right track. Our next goal will be to discover what e^(i*x) is. This is because to produce Euler’s identity, we need to come up with:

${ e }^{ ix }=\cos { \left( x \right) } +i\sin { \left( x \right) }$

To come up with the formula above, we will need the table below, because our latest e^(x) formula will have to be transformed. x will be turned into i*x.

Imaginary Numbers Exponentiated
$\sqrt { -1 } =i$
$i\cdot i={ i }^{ 2 }=-1$
${ i }^{ 3 }={ i }^{ 2 }\cdot i=-i$
${ i }^{ 4 }={ i }^{ 3 }\cdot i=-i\cdot i=-{ i }^{ 2 }=1$
${ i }^{ 5 }={ i }^{ 4 }\cdot i=i$
${ i }^{ 6 }={ i }^{ 5 }\cdot i=i\cdot i={ i }^{ 2 }=-1$
${ i }^{ 7 }={ i }^{ 6 }\cdot i=-i$
${ i }^{ 8 }={ i }^{ 7 }\cdot i=-i\cdot i=-{ i }^{ 2 }=1$
${ i }^{ 9 }={ i }^{ 8 }\cdot i=i$
${ i }^{ 10 }={ i }^{ 9 }\cdot i={ i }^{ 2 }=-1$
${ i }^{ 11 }={ i }^{ 10 }\cdot i=-i$

As we’ve got the table above, we can figure out what the formula e^(i*x) would look like:

Since:

This means that:

${ e }^{ ix }=\cos { \left( x \right) } +i\sin { \left( x \right) }$

And finally, when x=π:

${ e }^{ i\pi }=\cos { \left( \pi \right) } +i\sin { \left( \pi \right) } \\ \\ \therefore \quad { e }^{ i\pi }=-1\\ \\ \therefore \quad { e }^{ i\pi }+1=0$

This is because:

$\cos { \left( \pi \right) } =-1\\ \\ \sin { \left( \pi \right) } =0$

You have produced Euler’s identity from almost absolute scratch. Give yourself a pat on the back! 🙂

Related:

Deriving the Taylor Series from scratch