Find all complex numbers which make the following equations true: $ |z+1| =1 $ and $ |z^2+1| =1 $

Find all complex numbers which make the following equations true:

$$ |z+1| =1 $$

$$ |z^2+1| =1 $$

Solution:

If $ |z+1| =1 $ holds true, then

$$z+1 = 1.e^{i2n\pi}$$

$$z = 1.e^{i2n\pi}-1$$

$$z = 1.e^{i2n\pi}-1e^{i2m\pi}$$

If $ |z^2+1| =1 $ holds true, then

$$z^2+1 = 1.e^{i2p\pi}$$

$$z^2 = 1.e^{i2p\pi}-1$$

$$z^2 = 1.e^{i2p\pi}-1e^{i2q\pi}$$

1. How to proceed after this?
2. Am I supposed to do in this manner or break complex number z into real part x and imaginary part y and get two equations and thus solve for x and y?

Solution 1:

If you write $z$ as $a+bi$, then you get the system$$\left\{\begin{array}{l}(a+1)^2+b^2=1\\(a^2-b^2+1)^2+(-2ab)^2=1,\end{array}\right.$$which is equivalent to$$\left\{\begin{array}{l}a^2+b^2+2a=0\\2 a^2 b^2+a^4+2 a^2+b^4-2 b^2=0.\end{array}\right.$$From the first equation, you get that $b^2=-2a-a^2$, and if you replace $b^2$ by $-2a-a^2$ in the second equation, you get $8a^2+4a=0$. So, $a=0$ or $a=-\frac12$, and therefore the solutions of your system are $0$ and $-\frac12\pm\frac{\sqrt3}2i$.

Solution 2:

It is clear that $z=0$ is a solution. Hence we assume $z \ne 0$.

From $|z+1|^2=1$ we derive $ |z|^2= -(z+ \bar z) \quad (1)$.

From $|z^2+1|^2=1$ we derive $|z|^4= -(z^2+ \bar z^2) \quad (2)$.

Then $(1)$ gives: $|z|^4= z^2+ \bar z^2+2z \bar z=z ^2+ \bar z^2+2|z|^2$.

With $(2)$ we derive $|z|^4=-|z|^4+2|z|^2.$ Hence $|z|^4=|z|^2$. This gives $|z|=1.$

Use again $(1)$ to see that $Re(z)=-1/2.$

From $1=|z+1|^2=1/4 + (Im(z))^2$ we get $Im(z)= \pm \frac{\sqrt{3}}{2}.$