# 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.

• @ArchisWelankar No. As per triangle inequality, the maximum value of $|z_1 - z_2| = |z_1| + |z_2|$. – Ashish Gupta Feb 6 '16 at 18:20
• @Freelancer. For complex $z_1,z_2$ we always have $|z_1+z_2\leq |z_1|+|z_2|$. We have equality (with non-zero $z_1$) only when $z_2/z_1$ is a non-negative real. My point is that if you have $|a-b+c|\leq |a-b|+|c|\leq |a|+|b|+|c|$ it may not be possible to have both $a/b\geq 0$ and $b/c\geq 0$ at the same time. I have posted an A to the Q. – DanielWainfleet Feb 7 '16 at 5:24
• @Freelancer. I mean the complex number $a/b$ is a non-negative real. For non-zero $a,b$ this means that $|a|+|b|=|a+b|$ only when $a,b,$ and $0$ lie on a line in the complex plane with $0$ not lying between $a$ and $b.$ And there is an error in my previous comment. For $|a-b|=|a|+|b|$ to hold, we require $a/b\leq 0$. Draw a diagram showing $a,b,0$ and $a+b$.You will see why it is called the Triangle Inequality – DanielWainfleet Feb 7 '16 at 6:50
• @user254665 Also the OP is asking for the proof by using the triangular inequality...do you think it is not possible to prove this result just by using triangular inequality...?? – Freelancer Feb 7 '16 at 7:38
• Why do you think the triangle inequality is all that is needed to find a tight bound? Triangle inequality estimates are typically far from tight. – Eric Towers Feb 7 '16 at 9:36

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 $|V|=14$.

• I'm waiting to see whether someone will post a sophisticated 2-line proof. – DanielWainfleet Feb 7 '16 at 5:47
• Thank you for the solution, but I want to avoid using brute force since it tends to become long. Can you think of some way of doing this question using triangle inequality? – Ashish Gupta Feb 7 '16 at 8:17

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.
• I am simply amazed that how can there be so many solutions to a small looking question....well I guess ...I can't say much more than this is just Straight from the book !!... – Freelancer Feb 8 '16 at 12:36
• But I am really having problems understanding the third proof based on Geometry I understood that $$|(z-(\mathrm{i}+\mathrm{i}\sqrt{2}))|×|(z-(\mathrm{i}-\mathrm{i}\sqrt{2}))|$$ represents the product of distance of $z$ from the points $(0,1+\sqrt{2})$ and $(0,1-\sqrt{2})$ but how is this product always a constant ? Why can't it change ...I know that if this product is constant then it will represent a Cassini oval ...but how to prove that this product is definitely a constant..? Maybe a rough diagram/graph can help me ..can you please explain this part a little bit ? – Freelancer Feb 8 '16 at 12:43
• @Freelancer : The product is not a constant. For each positive constant, there is a set of points having that product of distances. This is similar to the level sets of $|z-\mathrm{i}|^2$, which is a family of concentric circles, one circle per constant. I can't make a better diagram than the one at the link I provided. – Eric Towers Feb 8 '16 at 13:21
• @Freelancer : Further, we don't ask that the solution be on one level set. We construct the circle of radius $3$, notice that it cuts across several level sets and ask: "These level sets increase in height as we move away from the foci. At which point on the circle have we moved the greatest distance from the foci?" – Eric Towers Feb 8 '16 at 13:29
• @Eric towers so basically we construct a sort of family of many Cassini ovals having (two fixed points that is foci ) from which distances are measured....and then we make a circle($|z|=3$) keeping these in mind...but what then..?? I have made this sort of approximate image ...without bothering much about real/imaginary axes ...is this right ??(the circle is represented by a thick red curve)..i.stack.imgur.com/HmWcv.jpg – Freelancer Feb 8 '16 at 15:02

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...

• Sorry ...if my handwriting is bad....i t was a little bit long so instead of spending hours in writing this here ...which takes a lot of time on a smartphone....I thought it would be better to post this here quickly...sorry for any inconvenience in reading the images .... – Freelancer Feb 7 '16 at 4:19
• In line 6 you assert that the LHS and RHS attain their maximum values with the same $x$. This is incorrect. Also it turns out that $|z^2-2 i z+1|$ attains its max at $z=-3 i$ which is when $|z^2+1|$ is minimized. – DanielWainfleet Feb 7 '16 at 5:42
• The error you've made is in the last line. If we take $|z^2 + 1 - 2iz|$ as $a$ and $|z+1|^2$ as $b$ (to make explanation easy). What you have is $a \le b + 6$ and $8 \le b \le 10$. Since the maximum possible value of $b$ is 10, you can say that $a \le 10 + 6$ for all values of $a$ and $b$. However, this in now way provides a lower bound for $a$. You might also argue that the minimum value of $b+6$ is $14$, so $a$ must always be $\le 14$, however, this is not true either (generally. For this question, it turns out to be), since $a$ and $b$ are both functions of $z$. – Ashish Gupta Feb 7 '16 at 9:08