Proving $|1+z|< \dfrac{1}{2} \implies |1+z^2|> 1$ If $z$ is a complex number and $|1+z|< \dfrac{1}{2}$ how can I prove that:


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*$|1+z^2|> 1$.

 A: There're different approaches to this problem, algebraic and geometric being the most evident. Let's talk about the algebraic one.
The first inequality describes a disc with the center in $z_0=-1$ with the radius $\frac 12$, i.e. $$D=\left\{z\in \Bbb C:z=-1+re^{i\theta},\,\theta\in[0,2\pi),\,r\in [0,1/2) \right\}.$$
Let's study the real part of $z^2$, $z\in D$. We write
$$\Re z^2 =\Re ( (-1+re^{i\theta})^2) = \Re (r^2e^{2i\theta}-2re^{i\theta}+1)=1-2r\cos\theta+r^2\cos 2\theta$$
$$=1-2r\cos\theta +r^2 (2\cos^2\theta-1)=2(r^2\cos^2\theta- r\cos\theta +1/4)+1/2-r^2$$
$$=2(r\cos\theta+1/2)^2+1/2-r^2>\frac 14.$$
As a consequence $\Re (1+z^2)>5/4$.
Thus, we can write
$$|1+z^2|\ge|\Re (1+z^2)|>\frac 54.$$
Note that we obtain a stronger inequality than we initially wanted.
A: Draw a disk of radius $1/2$ centered at $(-1,0) \in \mathbb C$. Since $|(-1) - z| = |1+z| < \frac 12$, $z$ lies within the disk. If you express $z = re^{\imath \theta}$, a little trig shows that $\frac{5\pi}{6} < \theta < \frac{7\pi}{6}$. Now square to get $z^2 = r^2 e^{\imath 2\theta}$, where $\frac{5\pi}{3} < 2\theta < \frac{7\pi}{3}$. In particular, $z^2$ lies in the right-half plane which forces $|(-1) - z^2| = |1+z^2| > 1$.
A: Hint: Try to prove that the circle
$$|1+z| < \frac{1}{2}$$
is included in side 
$$D:= \{ z=R(\cos(\theta)+i \sin(\theta))| \frac{3\pi}{4} < \theta <\frac{5\pi}{4} \} \,.$$
This helps since then, for all $z \in D$ we have $\mbox{Re}(z^2) >0$. 
You can try to prove this wither by showing that $(1+R\cos(\theta))^2+R^2\sin^2(\theta) < \frac{1}{4}$ implies $\cos(\theta)< \frac{-\sqrt{2}}{2}$ oor much simpler, by finding the complex equations $\theta=\theta_{1,2}$ of the tangents through $z=0$ to the circle $|1+z|=\frac{1}{2}$.
