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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

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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.. Jatin Yadav · 3 years, 11 months ago

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@Jatin Yadav 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 · 3 years, 11 months ago

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@Jimmy Kariznov oh! yes , you are right: Jatin Yadav · 3 years, 11 months ago

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