Assume that $| z + 1 | > 2$. Show that $|z^3 + 1| > 1$.
My try was: $$|z^3 + 1| = |z + 1| |z^2 - z + 1| > 2 |z^2 - z + 1| $$ but I'm stuck proving that $|z^2 - z + 1| > \frac 1 2$
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1A slight mistake, $z^3+1=(z+1)(z^2-z+1)$. – Dec 03 '15 at 13:04
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I can't do it nevertheless – Involutive Automorphism Dec 03 '15 at 18:02
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In these cases, I always try doing the substitution $z=a+bi$. It turns the problem into an algebraic one, which can sometimes be easier to solve. – Cataline Dec 03 '15 at 18:38
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Write $z = -1+ w$, so $|z+1|>2$ means $|w| > 2$. Then $z^3 + 1 = w^3 - 3 w^2 + 3 w = w (w^2 - 3 w + 3)$. The claim is that if $|w| > 2$, $|w^2 - 3 w + 3| > 1/2$. If $w = r \exp(i\theta)$, $$|w^2 - 3 w + 3|^2 = r^4 - 6 r^3 \cos(\theta) + (3 + 12 \cos(\theta)^2) r^2 - 18 r \cos(\theta)+ 9$$ Call that $F(r,\theta)$. We have $$ \dfrac{\partial F}{\partial \theta} = 6 r \sin(\theta) (r^2 - 4 r \cos(\theta) + 3) $$ Thus for any $r > 2$ the minimum of $F(r,\theta)$ must occur at one of the values $\theta = 0, \pi$, or $\pm \arccos((r^2+3)/(4r))$ (note that $0 < (r^2+3)/(4r) \le 1$ if $2 < r \le 3$). For $r>2$ we have $$\eqalign{F(r,0) &= (r^2 - 3 r + 3)^2 > 1\cr F(r,\pi) &= (r^2+3r+3)^2 > 169\cr F(r,\pm \arccos((r^2+3)/(4r))) &= \dfrac{(r^2-3)^2}{4} > \dfrac{1}{4}\cr}$$ Thus the minimum is greater than $1/4$, which establishes the claim.
Robert Israel
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