# Binary operation with complex number

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.

• Start by writing down the definitions of closure, associativity and identity. – Yves Daoust Nov 2 '18 at 11:26
• Are you sure it isn't $z_a ∘ z_b= |z_a|e^{i\theta(z_b)}$? The way it is written, without the "i" in the exponent, combining two complex numbers always gives a real number so there cannot be an "identity". – user247327 Nov 2 '18 at 11:27
• How do you define the angle of $0$? Meaning, what it $1 \circ 0$? – Ofek Gillon Nov 2 '18 at 11:28

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.
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
• So it is not a monoid because it does not satisfy the identity property? I am a bit confused about the last part of your proof, which $Θ_u = Θ_z$ is a contradiction, i'm not sure where it exists? – BlackSky Nov 2 '18 at 11:37