If $|1+z|<\frac12$, then $|1+z^2|>1$

Question

What is the best way to treat proofs like

Let $z$ be a complex number. If $|1+z|<\frac12$, then $|1+z^2|>1.$

I have tried first by taking $z=x+iy$ and substitute in the given modulus inequalities and then by working with $z\bar{z}$ and taking squares. Both approaches give a lot of calculations. I am wondering if there are more suitable approaches for proofs like this. I am feeling that a geometric proof is the best here but I am puzzled how to proceed.

I am interested in both algebraic and geonetric approaches. Thank you very much in advance for any help.

Answer 1

We have that

$$|1+z|=\frac12 \iff z=\frac12e^{i\theta}-1$$

then

$$1+z^2=\frac14e^{i2\theta}-e^{i\theta}+2$$

and finally

$$|1+z^2|^2=(1+z^2)(1+\bar z^2)=2\cos^2\theta-\frac92\cos\theta+\frac{65}{16}\ge \frac{25}{16}$$

---

Here is a graphical representation for the mapping $z\to z^2$

(credit Squaring the Circle by Steve Phelps)

---

As an alternative we can also use that

$$1+z^2=\frac14e^{i2\theta}-e^{i\theta}+2 \implies \Re(1+z^2)=\frac12\cos^2\theta-\cos\theta+\frac{7}{4}\ge \frac{5}{4}$$

Answer 2

This answer does not make use of the maximum principle. Let $u=z+1=x+iy.$ Then $$z^2+1=u^2-2u+2\\ =(x^2-y^2-2x+2)+(2xy-2y)i$$ Next $$x^2-y^2-2x+2=2x^2-2x+2-(x^2+y^2)\\ \ge 2x^2-2x+{7\over 4}$$ The minimal value of the last expression is attained at $x={1\over 2}$ and is equal ${5\over 4}.$ Summarizing the minimal value is attained for $x={1\over 2}$ and $y=0.$