How do you find the maximum value of $|z^2 - 2iz+1|$ given that $|z|=3$, using triangle inequality? 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.
 A: After playing with the triangle inequality for a while, we may realize that we are not going to arrive at the maximum without absurd ingenuity, so we consider other methods:


*

*Calculus I: stationary points:  Substitute $z = 3 \mathrm{e}^{\mathrm{i}\theta}$, find real and imaginary parts and construct the modulus as the sum of the squares of those parts, giving (simplified) $$\sqrt{2} \left( \sqrt{59 + 9 \cos(2 \theta) - 48 \sin(\theta)} \right) \text{.}$$  Differentiate this with respect to $\theta$, giving $$ -\frac{3\sqrt{2} \left( 8 \cos(\theta) + 3 \sin(2\theta) \right)}{\sqrt{59 + 9 \cos(2 \theta) - 48 \sin(\theta)}} \text{.}$$  Set this equal to zero and solve for $\theta$, giving $\pm \pi/2$ as locations of stationary points.  Evaluating the substituted polynomial at these two angles gives $-2$ and $-14$, so the maximum modulus of the polynomial on the circle of radius $3$ is $14$.

*Lagrange Multipliers:  Construct $|z^2 + 2\mathrm{i} z + 1|
 - \lambda(|z| - 3)$ then take derivatives with respect to $z$ and
$\lambda$, set those simultaneously equal to zero and solve.  You get
that $z = \pm 3\mathrm{i}$.  Plugging in again, we find the maximum
modulus is 14.

*Geometry:  This polynomial is $(z-(\mathrm{i}+\mathrm{i}\sqrt{2}))(z-(\mathrm{i}-\mathrm{i}\sqrt{2}))$.  Taking the modulus, we realize the level sets are collections of points whose product of distances from two given point (the roots just found) are fixed.  These level sets are Cassini ovals.  By symmetry, then, the maximum will be on the imaginary axis and it is no great challenge to realize it will be the one of $3\mathrm{i}$ and $-3\mathrm{i}$ that is farthest from the midpoint of the roots (which is $\mathrm{i}$).  Plugging $-3i$ back into the polynomial, we get that the maximum modulus is $14$, again.

A: 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 $|V|=14$. 
A: I am posting this as an answer in form of images because Images are not supported in the comments.


Now so basically as pointed out in the comments even though we have determined the range of $|z^2+1|$ and know that its Maximum value is $10$ still we can't determine the maximum/minimum value of$|z^2-2iz+1|$...
So even though putting this (10)in the equation $$|z^2-2iz+1|\le 10+6 $$ and again keeping (8) in the equation $$|z^2-2iz+1| \le 8+6$$... we can only say this much ..but then again by looking at these two equations
$$|z^2-2iz+1|\le 10+6$$ 
$$|z^2-2iz+1| \le 8+6$$ 
So after looking at this(the two simultaneous equations) I think we can definitely say that $|z^2-2iz+1|$ must be less than or equal to $14$ because say we have $|z^2-2iz+1|$ and we know it is less than $16$ but we also know it is less than $14$ and so theese two can be simultaneously true only in one situation... and so our final inequality is...
$$|z^2-2iz+1| \le 14$$ so this indeed gives the right answer $14$ but as pointed out in the comments by Ashish ..this is not the right method in general and only works in some circumstances such as these...so I am not so sure about this method because its is not in general...
