Question No $1$
Find the maximum value of $f(z)=|z^3-z+2|$ on the unit circle $|z|=1$
Solution Let $z:=\exp(\text{i}\theta)$ and $t:=\cos(\theta)$. Then, $$\begin{align}|z^3-z+2|^2&=\Big|\big(\cos(3\theta)-\cos(\theta)+2\big)^2+\text{i}\big(\sin(3\theta)-\sin(\theta)\big)\Big|^2 \\&=4\cos(3\theta)-2\cos(2\theta)-4\cos(\theta)+6 \\&=4(4t^3-3t)-2(2t^2-1)-4t+6 \\&=16t^3-4t^2-16t+8=:g(t)\,. \end{align}$$ Thus, we want to maximize $g(t)$ subject to $t\in[-1, 1]$. Note that $$g'(t)=48t^2-8t-16=8(6t^2-t-2)=8(2t+1)(3t-2)\,.$$ That is, optimizing points of $g(t)$ in $[-1, 1]$ are $t=-\dfrac12$, $t=\dfrac23$, and the boundary points $t\in\{-1, 1\}$.
Note that $g(-1)=4$, $g(1)=4$, $g\left(-\dfrac12\right)=13$, and $g\left(\dfrac23\right)=\dfrac{8}{27}$. We conclude that the minimum of $g(t)$ on $[-1,+1]$ is $\dfrac{8}{27}$, which is attained when $t=\dfrac23$, whereas the maximum of $g(t)$ on $[-1, 1]$ is $13$, which is attained when $t=-\dfrac12$. Translating this back to $\theta$, we see that $\theta=\dfrac{2\pi}{3}$ and $\theta=\dfrac{4\pi}{3}$ satisfy $\cos(\theta)=-\dfrac12$. Thus, $$z=\exp\left(\frac{2\pi\text{i}}{3}\right)=\dfrac{-1+\sqrt{3}\text{i}}{2}\text{ and }z=\exp\left(\frac{4\pi\text{i}}{3}\right)=\dfrac{-1-\sqrt{3}\text{i}}{2}$$ maximize $|z^3-z+2|$ for $z\in\mathbb{C}$ with $|z|=1$. The maximum value of $|z^3-z+2|$ for $z\in\mathbb{C}$ with $|z|=1$ is $\sqrt{13}$.
Similarly, the minimum value of $|z^3-z+2|$ for $z\in\mathbb{C}$ with $|z|=1$ is $\sqrt{\dfrac{8}{27}}=\dfrac{2\sqrt{2}}{3\sqrt{3}}$. The minimum is attained when $$z=\frac{2+\sqrt{5}\text{i}}{3}\text{ and }z=\frac{2-\sqrt{5}\text{i}}{3}\,.$$ (This happens when $\theta=\text{arccos}\left(\dfrac23\right)$ and when $\theta=2\pi-\text{arccos}\left(\dfrac{2}{3}\right)$.)
Doubt: Why can't i apply traingle inequality here? which is $||Z_1|-|Z_2||\leq |Z_1+Z_2| \leq |Z_1|+|Z_2|$
Question No $2$
Let $f(z)=2z^2-1.$Then what is the maximum value of $|f(z)|$ on the unit disc $D=\{z \in \mathbb C:|z|\leq 1\}$ ?
Solution$|f(z)|=|2z^2-1|\leq 2|z|^2+1\leq 2+1=3.$ So, is $3$ the maximum value of $|f(z)|?$ Can someone point me in the right direction? Thanks in advance for your time.
We get answer using triangle inequality here
Question No $3$
How do you find the maximum value of $|z^2 - 2iz+1|$ given that $|z|=3$, using triangle inequality?
Solution 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][1]: 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][2]: 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][3]. 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.
Here too we did not get maximum value using triangle inequality
Question No $4$
If $|Z-(3+4i)|=2$ then maximum value of $|Z|$ is
Solution Using Triangle ineqaulity $ ||Z|-|3+4i||\leq |Z-(3+4i)|$
$\implies$ $||Z|-|3+4i|| \leq 2$
$\implies$ $\implies$ $ -2 \leq|Z|-|3+4i| \leq 2$
$\implies$ $\implies$ $-2+5||Z| \leq 2+5$
$\implies max(|Z|)=7$
Here i got answer using triangle inequality
Final Doubt: I am a teacher my doubt is how to tell a class 12th students When to apply tringle inequality and when not?
Find the maximum value of $f(z)=|z^3-z+2|$ on the unit circle $|z|=1$
What is the maximum value of $|f(z)|$ on the unit disc $D=\{z \in \mathbb C:|z|\leq 1\}$?
How do you find the maximum value of $|z^2 - 2iz+1|$ given that $|z|=3$, using triangle inequality?
Regarding question 4.
I suggest drawing pictures whenever possible. (A student feels great when she can see the answer.) You got an upper bound, which you don't know is the maximum value. This question is really about the cartesian plane. You have a circle of radius 2 centered at (3,4). The maximum (and minimum) distance points lie on the line through the origin and the center of the circle. (Justification: the line is normal to the circle when it crosses.) Algebraically, the maximum distance of $7$ occurs at $$(3,4)+2\cdot{1\over5}(3,4)={7\over5}(3,4)$$