In what region is $|z^2 + 9|<1$ for complex $z$?

Question

In what region is $|z^2 + 9|<1$ for complex $z$?

I wrote this inequality $|z - 3i||z + 3i|<1$.

So I know they are asking for the set of points the product of whose distances from $3i$ and $-3i$ is less than 1.

I'm thinking a good idea is to first find the set of points the product of whose distances from $3i$ and $-3i$ equals 1.

I believe I've found these points lying on the imaginary axis : $\sqrt8i, -\sqrt8i, \sqrt10i, -\sqrt10i$.

But no idea how to find the others, and more generaly, no idea how to solve the problem.

Answer 1

We have $$1>|z^2+9|\geq|z|^2-9\to |z|^2<10$$ and $$1>|z^2+9|\geq9-|z|^2\to |z|^2>8$$ then the desired area lies in $2\sqrt{2}<|z|<\sqrt{10}$.

Answer 2

If $z=x+iy$, then $z^2=(x^2-y^2)+2xyi$. Now $\lvert z^2+9\rvert\lt1$ translates to $\lvert (x^2-y^2+9)+2xyi\rvert\lt1\implies (x^2-y^2+9)^2+(2xy)^2\lt1\implies x^4-2x^2y^2+y^4+18x^2-18y^2+y^4+81+4x^2y^2\lt1\implies x^4+y^4+18x^2-18y^2+81+2x^2y^2\lt1 $

@dxiv points out that this is the region enclosed by a quartic curve... or, a Cassini Oval, according to @David... @user 108128 has shown that the region lies in a certain annulus...

Answer 3

As already pointed out, there are two regions, roughly centred on $3i$ and $-3i$ respectively. Suppose $x + iy$ lies on the boundary of the first region. Then $$(x + iy)^2 = -9 + e^i\theta$$ for some $\theta$ ($0 \le \theta \lt 2\pi$). Equate real and imaginary parts: $$\cos\theta = x^2 - y^2 + 9,\quad\sin\theta = 2xy.$$ Multiply the first by $4y^2$ and substitute for $(2xy)^2$ from the second: $$4y^4 + 4(\cos\theta - 9)y^2 -\sin^2\theta = 0,$$ a quadratic in $y^2$. Taking the positive root, and recalling that $y$ is roughly $3$ rather than $-3$, we get $$y = \sqrt{9 - \cos\theta + \sqrt{82 - 18\cos\theta}\over2}.$$ From $\sin\theta\ = 2xy$ above we get $$x = {\rm sgn}(\sin\theta)\sqrt{-9 + \cos\theta + \sqrt{82 - 18\cos\theta}\over2}.$$ If we put $u = \cos(\theta/2)$ then the boundary is parametrized by $$x = {\rm sgn}(u)\sqrt{-5 + u^2 + \sqrt{25 - 9u^2}},\quad y=\sqrt{5 - u^2 + \sqrt{25 - 9u^2}}$$ for $-1 \le u \le 1$ (where $u = \pm1$ give the same point $(0, \sqrt8)$).