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Show that if $|z| = 3$, then

$$8 ≤ |3z^2 − 5z + 4i| ≤ 46$$

How do I go about proving this? Should I use the triangle inequality here?

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  • $\begingroup$ Yes, the triangle inequality works. $\endgroup$ – Mark Viola Aug 23 '18 at 2:59
  • $\begingroup$ Yes, you can work it out with triangle inequality. $\endgroup$ – Saucy O'Path Aug 23 '18 at 3:00
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    $\begingroup$ $|z_1|-|z_2|-|z_3|\leq|z_1+z_2+z_3|\leq|z_1|+|z_2|+|z_3|$ $\endgroup$ – Nosrati Aug 23 '18 at 3:18
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As pointed out by @Mark, @Saucy and @Nosrati in the comments, we can proe the statement using the Triangle Inequality .


Proof -

Claim -

$\vert z \vert = 3 \Rightarrow 8 \leq \vert 3.z^2 - 5.z + 4i \vert \leq 46$

Solution -

We have, by the triangle inequality -

$\vert z_1 \vert - \vert z_2 \vert - \vert z_3 \vert \leq \vert z_1 + z_2 + z_3 \vert \leq \vert z_1 \vert + \vert z_2 \vert + \vert z_3 \vert$

$\Rightarrow \vert 3.z^2 \vert - \vert (-5).z \vert - \vert 4i \vert \leq \vert 3.z^2 - 5.z + 4i \vert \leq \vert 3.z^2 \vert + \vert (-5).z \vert + \vert 4i \vert$

Since $\vert z \vert = 3 $, we have -

$3.3^2 -5.3 -4 \leq \vert 3.z^2 - 5.z + 4i \vert \leq 3.3^2+5.3+4$

$\Rightarrow 27-15-4 \leq \vert 3.z^2 - 5.z + 4i \vert \leq 27+15+4$

$\Rightarrow 8\leq \vert 3.z^2 - 5.z + 4i \vert \leq 46$

Q.E.D.

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