8

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.

gebruiker
  • 6,154
Ashish Gupta
  • 1,633
  • It should be $-2|i||z|$ not $+$ – Archis Welankar Feb 06 '16 at 18:17
  • 7
    @ArchisWelankar No. As per triangle inequality, the maximum value of $|z_1 - z_2| = |z_1| + |z_2|$. – Ashish Gupta Feb 06 '16 at 18:20
  • The trouble with this approach is that you are using 2 inequalities that cannot both be equalities.E.g.,for the 2nd attempt $|(z-i)^2+2|\leq |z-i|^2+2\leq (|z|+|i|)^2+2\leq 18.$ But $|(z-i)^2+2|=|z-i|^2+2 $ only when $(z-i)^2$ is a non-negative real (and hence $z\ne -3 i$,) while $|z-i|^2+2=(|z|+|i|)^2$ only when $z =-k^2 i$ for some real $k$. A different approach is needed. – DanielWainfleet Feb 06 '16 at 22:42
  • @user254665 so basically you are saying that there is a condition (which you have given in the above comment ) for the inequality $$|z_1+z_2|\le|z_1|+|z_2|$$ to be true right..??...or I am misinterpreting this...? – Freelancer Feb 07 '16 at 04:55
  • 1
    @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 07 '16 at 05:24
  • @user254665 Do you mean that $\frac{a}{b}>= 0$ or both a,b should have value greater than $0$.. – Freelancer Feb 07 '16 at 06:25
  • 1
    @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 07 '16 at 06:50
  • @user254665 thanks ...I see now why the equality used is wrong...thanks for the explanation.... This question http://math.stackexchange.com/questions/652303/equality-of-triangle-inequality also helped me.. – Freelancer Feb 07 '16 at 07:35
  • @user254665 But then ...I don't see any such problem in my answer....can you point out my mistake in the answer below...I have edited it now.... – Freelancer Feb 07 '16 at 07:37
  • 1
    @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 07 '16 at 07:38
  • 1
    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 07 '16 at 09:36
  • Well..@Eric towers can you suggest some other methods of doing this..? Other than the brute force method used in the answer below....? – Freelancer Feb 07 '16 at 09:38

3 Answers3

5

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.
Eric Towers
  • 67,037
  • 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 08 '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 08 '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 08 '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 08 '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)..http://i.stack.imgur.com/HmWcv.jpg – Freelancer Feb 08 '16 at 15:02
  • @Eric towers And since we wanted to maximize $$|(z-(\mathrm{i}+\mathrm{i}\sqrt{2}))|×|(z-(\mathrm{i}-\mathrm{i}\sqrt{2}))|$$ so we wanted the product of the distances to be maximum and since as we move far from the foci of the Cassini ovals this product also maximises such as in the figure given above so we can say that this happens at $z=-3i,+3i$ ...right....?? – Freelancer Feb 08 '16 at 15:04
  • @Eric towers But then this also gives two answers that is (+3i,-3i ) why this discrepancy in the answer...?? Sure we can plug this value in The function to check for ourselves but still...is there any other way of determining which answer is wrong out of +3i,-3i other than putting the value ?? – Freelancer Feb 08 '16 at 15:09
  • @EricTowers Thanks for the answer. I love the Lagrange Multiplier method, but fail to understand two things about it. First, why is the first expression $z^2+2iz+1$? Shouldn't it be $|z|^2+2i|z|+1$? Secondly, if not, how do you take the derivative wrt $z$? If you treat $|z|$ as a constant, the derivative gives you $z=−i$. – Ashish Gupta Feb 14 '16 at 17:55
  • @Freelancer: The imgur diagram you made has some correct parts -- the level sets of the Cassii ovals are right. However, putting the center of the circle at the midpoint between the foci is entirely misleading. The center of the circle should be much closer to one of the foci. (The foci are at $1.4\dots \mathrm{i}$ and $-0.4\dots \mathrm{i}$, the center of the circle is at $0$.) Your schematic indicates a false left-right symmetry, which causes you to believe that $3\mathrm{i}$ and $-3\mathrm{i}$ are on the same level set. They are not. – Eric Towers Feb 14 '16 at 19:43
  • @Freelancer : My comments were coincident with StackExchange's downtime today, or I could have made the edits to my comment: "Cassini" ovals and "The foci are at $2.4\dots \mathrm{i}$ and ...". – Eric Towers Feb 14 '16 at 19:56
  • @AshishGupta : Oops. Dropped the modulus bars from my objective function... Fixed. This is not the same as using the triangle inequality to move the bars in to the $z$s. Recall that $|f|^2 = f \overline{f}$ and you can then be careful about which branch of $\sqrt{\phantom{z}}$ you want and then use the chain rule to differentiate. – Eric Towers Feb 14 '16 at 20:02
  • @EricTowers I know it isn't. But somehow, I found the use of Lagrange's Multipliers even better than Triangle Inequality. Can you please elaborate on the method of differentiation in the answer/here in the comments? I still don't understand how you'd differentiate it and get the answer. – Ashish Gupta Feb 15 '16 at 10:35
  • @AshishGupta : I could ... But instead, I'll use the fact that we're both using a large repository of math questions and answers to point out that this is not the first time your question has been asked (and answered). See this MSE question (and answer), and recall that the derivative is linear and polynomials are holomorphic. – Eric Towers Feb 15 '16 at 13:07
4

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 $|V|=14$.

DanielWainfleet
  • 57,985
  • 4
    I'm waiting to see whether someone will post a sophisticated 2-line proof. – DanielWainfleet Feb 07 '16 at 05:47
  • 1
    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 07 '16 at 08:17
0

I am posting this as an answer in form of images because Images are not supported in the comments. Image 1 Image 2 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...

Freelancer
  • 954
  • 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 07 '16 at 04:19
  • 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 07 '16 at 05:42
  • 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 07 '16 at 09:08