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

*

*How to proceed after this?

*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?

 A: 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 $-\frac12\pm\frac{\sqrt3}2i$.
A: 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 $Im(z)= \pm \frac{\sqrt{3}}{2}.$
A: If we interpret the absolute-value expressions as descriptions of loci of points at specified distances from particular locations, then
$   |z + 1| \ = \ 1 \ \ $ is the set of points at unit distance from $ \ z \ = \ -1 \ \ $ [the unit circle centered at that point] and
$ |z^2 + 1| \ = \ |(z + i)·(z - i)| \ = \ 1 \  \ $ is the set of points for which the product of the distances of each point from $ \ z \ = \ i \ $ and $ \ z \ = \ -i \ $ is a constant $ \ 1 \ $ [the "unit" lemniscate with foci at those points].
Solving for the intersections of these two curves using their Cartesian expressions becomes a bit complicated.  We might instead convert these to "polar form":
$ (x + 1)^2 + y^2 \ = \ 1 \ \rightarrow \ x^2 + y^2 + 2x + 1 \ = \ 1 \ \rightarrow \ r \ = \ -2·\cos \theta \ \ ; $
$ [ \ x^2 + (y + 1)^2 \ ] · [ \ x^2 + (y - 1)^2 \ ] \ = \ 1 \ \rightarrow \  x^4 + 2x^2y^2 + y^4 \ = \ 2·(y^2 - x^2) $ $ \rightarrow \ r^2 \ = \ -2·\cos(2 \theta) \ \ . $
Equating these polar curve equations will give us information about intersections other than the origin (which must be checked separately):
$$ -2·\cos(2 \theta) \ \ = \ \ ( \ -2·\cos \theta \ )^2 \ \ \Rightarrow \ \ -2 \ · \ ( \ 2·\cos^2 \theta \ - \ 1 \ ) \ \ = \ \ 4  ·  \cos^2 \theta $$
$$ \Rightarrow \ \ 8  ·  \cos^2 \theta \ \ = \ \ 2 \ \ \Rightarrow \ \ \cos \theta \ \ = \ \ \pm \ \frac12 \ \ . $$
The solution $ \cos \theta \   =   \ + \frac12 \ $ is "spurious", since it cannot be used in the equation $ \ r \ = \ -2·\cos \theta \ $ (alternatively, we can read the "negative radii" as duplicating the results for the positive-radius solutions).  For $ \cos \theta \ = \ - \frac12 \ \ , $ we have $ \ \theta \ = \ \frac{2 \pi}{3} \ \ , \ \ \frac{4 \pi}{3} \ \ $ and $ \ r \ = \ 1 \ $ on the circle.  For the lemniscate, we find $ \ 2\theta \ = \ \frac{4 \pi}{3} \ \ , \ \ \frac{8 \pi}{3} \ = \ \frac{2 \pi}{3} \ \ , $ so we again have $ \ r \ = \ 1 \ \ . $  The "off-origin" solutions are thus
$$ z_1 \ \ = \ \ e^{ \ i·2 \pi/3} \ \ = \ \ -\frac12 + i·\frac{\sqrt3}{2} \ \ \ , \ \ \ z_2 \ \ = \ \ e^{ \ i·4 \pi/3} \ \ = \ \ -\frac12 - i·\frac{\sqrt3}{2} \ \ = \ \ \overline{z_1} \ \ . $$  [It might be mentioned that these are the two complex cube-roots of unity.]
The origin is also an intersection point of these curves, since $ \ r \ = \ 0 \ $ and $ \ r^2 \ = \ 0 \ $ are possible values for the two curve functions, although they occur for different values of $ \ \theta \ $ for each curve.  So the third (trivial?) solution for the pair of equations is $ \ z \ = \ 0 \ \ \ . $

