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How would you prove that if $|z| \geq3$ then $|z^{5}+135| \geq 108$ where $z \in \Bbb C$

by De Moivre's theorem i can write $z^5$ as $r^5(\cos(5\phi)+i\sin(5\phi)$ and so $|z^{5}+135|=\sqrt{(r^5(\cos(5\phi)+135)^2+\sin^2(5\phi)}$ but at this point im stuck on how to use the fact that $r \geq 3$ to show the conclusion.

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  • $\begingroup$ I would suggest the "visual approach". The map $z\mapsto z^5$ brings the region $\|z\|\geq 3$ (the complex plane with a hole enclosing the origin) into the region $\|z\|\geq 243$ (the complex plane with a larger hole around the origin). Then the map $z\mapsto z+135$ brings the region $\|z\|\geq 243$ into the region $\|z-135\|\geq 243$ (the complex plane with a large hole enclosing the origin, off-centered). The hole contains the ball $\|z\|\leq 243-135=108$, hence every point of the region $\|z-135\|\geq 243$ has a distance from the origin which is at least $108$. $\endgroup$ – Jack D'Aurizio Nov 18 '17 at 0:13
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Triangle inequality implies that $|z^{5}+135|\geq|z^{5}|-135=|z|^{5}-135\geq 3^{5}-135=108$.

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  • $\begingroup$ How does $|z^5+135| \le |z^5|+|135|$ imply that $|z^5+135| \ge |z^5|-135$? $\endgroup$ – Max Li Nov 18 '17 at 20:39
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    $\begingroup$ We need the other way: $|z^{5}|=|z^{5}+135-135|\leq|z^{5}+135|+|-135|=|z^{5}+135|+135$. $\endgroup$ – user284331 Nov 18 '17 at 20:43
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More intuitively, $|z^5+135|$ is largest when $z^5$ points in the same direction as $135$ and smallest when it points in the opposite direction of $135$.

By direction I mean the vector from the origin to the point in the plane. Now since $135$ is a positive real, the quantity in question is largest when $z^5$ is a positive real and smallest when it is a negative real.

Since $|z|=3$, the values $\pm3^5\equiv \pm 243$ are the ones we are talking about. So the largest value of the quantity is $$|243 +135|=|378|=378$$ and the smallest is $$|-243+135| = |-108| = \boxed{108}$$

Note that there are several values of $z$ for which $z^5$ is a positive (or negative) real.

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  • $\begingroup$ could you also use the fact that $-1 \leq \cos(5\phi) \leq 1$ and similairly for $\sin(5\phi)$ $\endgroup$ – Skrrrrrtttt Nov 17 '17 at 21:52

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