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.
On
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.
Triangle inequality implies that $|z^{5}+135|\geq|z^{5}|-135=|z|^{5}-135\geq 3^{5}-135=108$.