# How do I determine all $z\in \mathbb{C}$ which satisfy the equation $\left | z^2-1 \right |< r$?

Let $r>0$. How do I determine all $z\in \mathbb{C}$ which satisfy the equation $\left | z^2-1 \right |< r$? How do I sketch and determine the values of r for which the set M ${z\in \mathbb{C}: \left | z^2-1 \right |< r}$ is related?

It’s clear that $z$ satisfies the equation $z^2-1=de^{i\theta}$ for some $d <r$. Solving, we get $z=\pm\sqrt {1+de^{i\theta}}$ We want to write this in the form $z=\pm ae^{it}$ Thus $a^2e^{i2t}=1+d\cos (\theta)+id\sin (\theta)=\sqrt {(1+d\cos\theta)^2+d^2\sin^2\theta}\exp [i\tan^{-1}(\frac {d\sin\theta}{1+d\cos\theta})]$. Therefore $z$ is $z=\pm (1+d^2+2d\cos\theta)^{1/4}\exp [\frac {i}{2}\tan^{-1}(\frac {d\sin\theta}{1+d\cos\theta})]$ With $d <r$ and $0\leq\theta\leq2\pi$.

I have tried this but it's not correct. Can someone help me?

• Putting $z=x+iy$, we need the region $(x^2-y^2)^2 + 4x^2y^2 < r^2$. This is the region bounded by a Lemniscate (figure eight). – user348749 Mar 21 '17 at 10:49
• @Muralidharan It should be $(x^2-y^2-1)^2+4x^2y^2<r^2$ and it is a lemniscate only for $r=1$. – egreg Mar 21 '17 at 11:26

Given $\theta\in[0,2\pi)$, we want to find for what $d>0$ the relation $$|d^2e^{2i\theta}-1|<r$$ which is equivalent to $$|d^2e^{2i\theta}-1|^2<r^2$$ The modulus squared on the left-hand side can be written $$(d^2e^{2i\theta}-1)(d^2e^{-2i\theta}-1)= d^4-2d^2\cos2\theta+1$$ so we have the inequality $$d^4-2d^2\cos2\theta+1-r^2<0$$ Seeing this as a quadratic in $d^2$, the discriminant should be positive, so $r^2-\sin^22\theta>0$, hence $-r<\sin2\theta<r$.

Let's distinguish the cases $r>1$, $r=1$ and $0<r<1$.

In the case $r<1$ there is no condition on $\theta$ and the solutions in $d^2$ are $$0<d^2<\cos2\theta+\sqrt{r^2-\sin^22\theta}$$ Also $z=0$ satisfies the condition.

In the case $r=1$, we get $d^2<2\cos2\theta$; in particular, we must have $\cos2\theta>0$ in order for $d$ to exist.

In the case $0<r<1$, we must have $-r<\sin\theta<r$ and also $$\cos2\theta-\sqrt{r^2-\sin^22\theta}<d^2<\cos2\theta+\sqrt{r^2-\sin^22\theta}$$

Here are the graphs for the three cases. In all three cases, the solutions are the open regions bounded by the line(s).

## $r=1/\sqrt{2}$

I add also the case $r=3$; it would be interesting to see from what value of $r$ we start getting the oval shape.