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?
