Category Archives: Imaginary Numbers

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…

z_1 divided by z_2

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)

arg(z_1*z_2)=arg(z_1)+arg(z_2) Proof

In this post I’ll be proving why:

arg\left( { z }_{ 1 }{ z }_{ 2 } \right) =arg\left( { z }_{ 1 } \right) +arg\left( { z }_{ 2 } \right)

Let’s say that:

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

And also that:

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

This would imply that:

arg\left( { z }_{ 1 } \right) ={ \theta }_{ 1 }

arg\left( { z }_{ 2 } \right) ={ \theta }_{ 2 }

Now if we multiply { z }_{ 1 } and { z }_{ 2 } together, we get:

z_1 multiplied by z_2

Which is thanks to what we know about trigonometric identities.

As we can see above, we’ve formed another complex number:

{ z }_{ 1 }{ z }_{ 2 }={ r }_{ 1 }{ { r }_{ 2 }\left( \cos { \left( { \theta }_{ 1 }+{ \theta }_{ 2 } \right) +i\sin { \left( { \theta }_{ 1 }+{ \theta }_{ 2 } \right) } } \right) }

And this is in the form of:

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

And because of the rules of complex numbers, we can say that:

arg\left( { z }_{ 1 }{ z }_{ 2 } \right) \\ \\ ={ { \theta } }_{ 1 }+{ { \theta } }_{ 2 }\\ \\ =arg\left( { z }_{ 1 } \right) +arg\left( { z }_{ 2 } \right)

Hence, we have our 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…