Let us consider the set of complex numbers and the binary operation $\circ$ defined by
$z_a\circ z_b=|z_a|e^{\Theta(z_b)}$,
where $\Theta(z_b)$ is the argument of the complex number $z_b$.
Explain whether the set of complex numbers together with the binary operation $\circ$ forms a monoid.
I know I need to prove whether it satisfies closure, associativity and identity.
I'm not sure how to start with showing this, I've tried to put it into polar form which I get $z_a\circ z_b=|z_a|(\cos \Theta(z_b)+i\sin \Theta(z_b))$ but I have no idea what to do next.
Any help will be appreciated, thanks.
Well, closure is trivial because this operation will return a complex number.
About associativity, let there be 3 complex numbers $z_1,z_2,z_3$ and we'll try to prove that $$(z_1 \circ z_2) \circ z_3 = z_1 \circ (z_2 \circ z_3) $$ The left side will be $$|z_1| e^{i \Theta_2 } \circ z_3 = |z_1| e^{i \Theta_3} $$ and the right side will be $$ z_1 \circ |z_2| e^{i\Theta_3} = |z_1| e^{i \Theta_3} $$ which is the same, so associativity holds.
Now, about the unity part, I'll try to prove that there is no such thing in this set.
Let's assume that there is a unity, $u = |u| e^{i \Theta_u} $. For any $z$, $z\circ u = |z| e^{i \Theta_u} $ which needs to be equal to $z=|z| e^{i\Theta_z}$, meaning that $$ \Theta_u = \Theta_z$$ For every $z$, meaning that by taking any two complex numbers with different angles will contradict the assumption. QED