Problem:
How do you find the maximum value of $|z^2 – 2iz+1|$ given that $|z|=3$, using triangle inequality?
My attempt:
$$|z^2 – 2iz+1|\le|z|^2+2|i||z|+1$$
$$\implies |z^2 – 2iz+1|\le16$$
However, this does not provide a strict upper bound on the inequality, where the equality holds.
I also tried writing it as:
$$|(z-i)^2 + 2| <= |(z-i)|^2 + 2$$
This last equation does suggest that the maximum value occurs at $-3i$, however, provides an even higher upper bound of $18$.
Wolfram Alpha gives the answer as $14$, and it occurs as $-3i$. I know that the equality only holds when all the complex numbers are collinear, but that has not helped me with this question.
Best Answer
Brute force. Let $z=3(c+i s)$ where $c=\cos t$ and $s=\sin t$ with $t\in R$. Let $V= z^2-2 i z+1.$ Then$$ V=(z-i)^2+2=(3 c +i(3 s-1)^2+2=9 c^2-(3 s-1)^2 +6 i c(3 s-1)+2=$$ $$=9c^2-9 s^2+6 s+1+i(18 c s-6c)=(9 c_2+6 s+1) +i(9 s_2-6 c)$$ where $c_2=c^2-s^2=\cos 2 t$ and $s_2 =2 c s=\sin 2 t.$....... So we have$$|V|^2=(81 c_2^2+36 s^2+1+108 c_2 s+18 c_2+12 s)+(81 s_2^2-108 s_2 c+36 c^2).$$ Now $81 c_2^2+81 s_2^2=81$ and $36 s^2+36 c^2=36,$ while $108(c_2 s-s_2 c)=108 (\cos 2 t \sin t-\sin 2 t\cos t=108(\sin (t-2 t)=-108 s.$.... So after simplifying we have $$|V|^2=118-96 s+18 c_2=118-96s +18(1-2 s^2)=136-96s -36 s^2$$( because $c_2=\cos 2 t= 1-2 \sin^2 t=1-2 s^2$.)..... Since $-1\leq s\leq 1$ the problem is to find the maximum value of $136-96 s-36 s^2$ for $s\in [-1,1]$, which is easily seen to be $196$, attained when $s=-1$. So $|V|^2\leq 196=14^2$.... When $s=-1$ we have $z= -3 i$ and $V=-14$ and $|V|=14$.