Complex Inequality

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

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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

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

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

Hence, i think it should be $2$ rather..

jatin yadav - 7 years, 9 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 $|z+1|^2 - 3|z| > 4 - 3 = 1$ .

Jimmy Kariznov - 7 years, 9 months ago

oh! yes , you are right:

jatin yadav - 7 years, 9 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$