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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.

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  • $\begingroup$ Start by writing down the definitions of closure, associativity and identity. $\endgroup$ – Yves Daoust Nov 2 '18 at 11:26
  • $\begingroup$ 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". $\endgroup$ – user247327 Nov 2 '18 at 11:27
  • $\begingroup$ How do you define the angle of $0$? Meaning, what it $ 1 \circ 0$? $\endgroup$ – Ofek Gillon Nov 2 '18 at 11:28
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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

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  • $\begingroup$ 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? $\endgroup$ – BlackSky Nov 2 '18 at 11:37
  • $\begingroup$ If there is an identity with a well defined angle, it can't be equal to the angle of any other complex number $\endgroup$ – Ofek Gillon Nov 2 '18 at 11:43
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    $\begingroup$ I think I get it now, it comes from the concept that e is unique so the angles cannot be equaled. Thank you! $\endgroup$ – BlackSky Nov 2 '18 at 11:48

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