×

# Complex Inequality

$$z$$ are complex number such that $$|z+1|> 2$$ , prove that $$|z^3+1| > 1$$

Note by Pebrudal Zanu
3 years, 11 months ago

Sort by:

$$|z| + 1 \geq | z + 1| > 2 , \Rightarrow |z| > 1$$

$$|z^3 + 1| = |z + 1||z^2 - z + 1| = |z + 1|| (z + 1)^2 - 3z| >|z + 1|( |z + 1|^2 - 3|z|) > 2$$.

( Using $$|z + 1| > 2$$ and $$|z| > 1$$).

Hence, i think it should be $$2$$ rather.. · 3 years, 11 months ago

Since $$|z+1| > 2$$, we definitely have $$|z+1|^2 > 4$$. However, $$|z| > 1$$ yields $$-3|z| < -3$$ instead of $$-3|z| > -3$$. So, we cannot immediately conclude that $$|z+1|^2 - 3|z| > 4 - 3 = 1$$. · 3 years, 11 months ago