Why $|z^5|-2^5=0$ has $\infty$ solutions in $\mathbb{C}$?

Question

Given $z=x+iy$ a complex number, I can't understand why $|z^5|-2^5=0$ has infinite solutions in $\mathbb{C}$.

Answer 1

Let's start by showing that $|z|-2=0$ has infinitely many solutions. Rewrite this as $|z|=2$. This equation is solved if $z$ has magnitude $2$; in other words, if it lies on the circle $x^{2}+y^{2}=2^{2}$. Clearly there are infinitely many points on this circle, so there are infinitely many such complex numbers.

Now, we have shown there are infinitely many complex numbers with magnitude $2$. Since $|z^{5}|=|z|^{5}$, each of the above numbers satisfies $|z^{5}|=|z|^{5}=2^{5}$

Answer 2

First off, note that $|z^5| = |z|^5$, so we can just drop the two $5$s, and ask why $|z| = 2$ has infinitely many roots in $\mathbb{C}$ (because it does, and all of these are necessarily solutions to your equation). But this is precisely what $z \mapsto |z|$ does: it takes each circle centred on the origin, and maps all points of that circle to the unique point on the positive real half-line. Thus, the infinitely many solutions that you speak of are exactly the infinitely many points on the circle of radius $2$.

Answer 3

The stated equation is equivalant to $|z|=2$. Now there are infinitely many $z$ that fullfill this:

$$z=2e^{i\phi}, 0\le \phi < 2\pi$$

would be one way to describe them, or

$$z=x\pm i\sqrt{4-x^2}, -2 \le x \le 2$$

would be another. Using the usual visualization of the Gaussian plane, those are the complex $z$ that lie on the circle of radious $2$ around the origin.