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\(z\) are complex number such that \(|z+1|> 2\) , prove that \(|z^3+1| > 1\)

Note by Pebrudal Zanu 4 years, 10 months ago

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\(|z| + 1 \geq | z + 1| > 2 , \Rightarrow |z| > 1\)

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

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

Hence, i think it should be \(2\) rather..

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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 \(|z+1|^2 - 3|z| > 4 - 3 = 1\).

oh! yes , you are right:

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`*italics*`

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italics`**bold**`

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`\frac{2}{3}`

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`\sin \theta`

`\boxed{123}`

## Comments

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TopNewest\(|z| + 1 \geq | z + 1| > 2 , \Rightarrow |z| > 1\)

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

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

Hence, i think it should be \(2\) rather..

Log in to reply

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 \(|z+1|^2 - 3|z| > 4 - 3 = 1\).

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oh! yes , you are right:

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