Suppose $|z|\ge 2$, Prove $|z^8+135|\ge121$.

Question

Suppose $|z|\ge 2$, Prove $|z^8+135|\ge121$.

My work: $|z^8+135|=\sqrt{(z^8+135)(\bar{z}^8+135)}=\sqrt{|z|^{16}+135(z^8+\bar{z}^8)+135^2}\ge\sqrt{2^{16}+135(z^8+\bar{z}^8)+135}$

For the last term, I don't know how to deal with $135(z^8+\bar{z}^8)$ to make an estimation. Any hints would be helpful.

Answer 1

By the reverse triangle inequality: $$|z^8+135|\geq |z^8|-135\geq 2^8-135=256-135=121.$$

Answer 2

$|z^8+135|$ is smallest when $z^8$ is a negative real number of largest possible magnitude. This is the case for $$z = 2e^{\frac{i\pi}8} = \sqrt{2+\sqrt{2}} + i\sqrt{2-\sqrt{2}}$$ so $$|z^8+135| = |-256+135| = |-121| = 121$$

Answer 3

By the triangle inequality: $$|z^8+135|=|z^8+135|+|-135|-135\geq|z^8+135-135|-135=$$ $$=|z|^8-135\geq2^8-135=121.$$

Answer 4

Alternatively: $$z=a+bi=r(\cos t+i\sin t) \Rightarrow |z|=r\ge 2;\\ z^8=r^8(\cos (8t)+i\sin (8t)) \Rightarrow \\ |z^8+135|=\sqrt{(r^8\cos (8t)+135)^2 +(r^8\sin (8t))^2} \ge 121 \iff \\ r^{16}+270r^8\cos (8t)+135^2\ge 121^2 \iff \\ r^{16}+270r^8(2\cos^2 (4t)-1)+135^2\ge 121^2 \iff \\ r^{16}-270r^8+135^2+540r^8\cos^2 (4t)\ge 121^2 \iff \\ \underbrace{(r^8-135)^2}_{\ge (2^8-135)^2=121^2}+\underbrace{540r^8\cos^2 (4t)}_{\ge 0}\ge 121^2.$$