Maximum and minimum modulus of a polynomial function $f(z)=z^{10}+1000i$ Let $f(z)=z^{10}+1000i$. Then on the annulus $1 \leq \mid z \mid \leq 2$, what are the maximum modulus and minimum modulus of $f(z)$?

From geometric interpretation, I think the maximum modulus and minimum modulus are all achieved on the circle $\mid z \mid =2$. Also, each value is $2024$ and $24$ respectively (when $z^{10}=1024i$ and $ -1024i$). Is my result correct? Could anyone please give me a feedback?
Hint
$$|z_1|-|z_2|\leq |z_1+z_2| \leq |z_1|+|z_2|$$

Equalities can be achieved only when $z_1=\lambda z_2$ or $z_2=\lambda z_1$ for some $\lambda \in \mathbb R$, with positive $\lambda$ achieving the sum and the negative achieving the difference.
You can convert it into a easy trigonometric problem. 
Write $z$=$re^{i\theta}$ . Now substituting z in required expression we have
 $f(z)$ = $r^{10}e^{10i\theta}+1000i$ = $r^{10}\cos{10\theta} + ir^{10}\sin{10\theta}+1000i$ = $r^{10}\cos{10\theta}$ + i($r^{10}\sin{10\theta}$+$1000$) 

Modulus of $a+ib$ is equal to $\sqrt{a^2+b^2}$ . Using this we have,
$\mid f(z) \mid = \sqrt{r^{100}\cos^2{10\theta}+r^{100}\sin^2{10\theta}+1000^2+2.1000.r^{10}\sin{10\theta}}$  
$\mid f(z) \mid = \sqrt{r^{100}+1000^2+2.1000.r^{10}\sin{10\theta}}$


For maximum modulus take r=2 and $10\theta = \pi/2$ 
For minimum modulus $10\theta = -\pi/2$ and take r such that it minimises $r^{100}-2.1000.r^{10}$