# 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: 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.