0
$\begingroup$

I derived a problem to the equation $$1=zz*$$ where $z*$ is the complex conjugate of $z$, and apparently it admits an infinite amount of solutions in $\mathbb C$ but I do not understand why so?

I know that the number of solutions in the complex space is always equal to the highest degree of the $p(z)$ polynomial i.e for $$z^2= 4$$ you would have 2 solutions (in this case the complex solution and its conjugate)

But how could I proceed here to find the number of solutions in $\mathbb C$?

  • 0
    $z^2=4$ has two real solutions which are their own conjugates. – 2017-01-11

2 Answers 2

3

$$zz^{*}$$ Is not a polynomial equation, so it does not have only $2$ or finite solutions. Instead, set $z=a+bi$. Now note $$zz^{*}=a^2+b^2=1$$

Thus for any $\theta \in \mathbb{R}$ if $z=\cos \theta +i \sin \theta$ then $$zz^{*}=1$$

  • 2
    Indeed, this could serve as a proof of the fact that $z^*$ is not a polynomial function in $z$. (You could also prove it by noticing that it's not complex differentiable, I guess.) – 2017-01-11
1

Recall that $zz^* = |z|^2$, that is, the product of a complex number with its conjugate is equal to the square of the norm. So the solutions to your equation are exactly the complex numbers of norm 1. This would mean that all complex numbers on the unit circle are solutions, so we have infinitely many solutions.