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An inequality involving complex numbers

March 19, 2024 beni22sof Leave a comment

Consider ${n\geq 1}$ and ${n}$ complex numbers ${z_1,...,z_n \in \Bbb C}$ . Show that

$\displaystyle \sum_{k =1}^n |z_k||z-z_k|\geq \sum_{k=1}^n |z_k|^2, \text{ for every } z \in \Bbb{C},$

if and only if ${z_1+...+z_n = 0}$ .

Proposed by Dan Stefan Marinescu in the Romanian Mathematical Gazette

Solution: For someone familiar with optimality conditions in multi-variable calculus this is straightforward. Notice that

$\displaystyle f(z) = \sum_{k =1}^n |z_k||z-z_k|$

is a convex function (linear combination of distances in the plane). The inequality is equivalent to ${f(z) \geq f(0)}$ , which means that ${0}$ is the global minimum of the function.

For a convex, differentiable function global minimality is equivalent to verifying first order optimality conditions. Denoting ${z = x+iy}$ , ${z_k = x_k+iy_k}$ the partial derivatives of the function ${f}$ with respect to ${x,y}$ are

$\displaystyle \frac{\partial f}{\partial x}(x,y) = \sum_{k=1}^n \sqrt{x_k^2+y_k^2}\frac{x-x_k}{\sqrt{(x-x_k)^2+(y-y_k)^2}},$

$\displaystyle \frac{\partial f}{\partial y}(x,y) = \sum_{k=1}^n \sqrt{x_k^2+y_k^2}\frac{y-y_k}{\sqrt{(x-x_k)^2+(y-y_k)^2}}.$

If ${\frac{\partial f}{\partial x}(0,0) = \frac{\partial f}{\partial y}(0,0)}$ then ${\sum x_k=\sum y_k = 0}$ and the conclusion follows. The converse also holds, obviously.

Since this problem was proposed for 10th grade, let’s use some simpler arguments to arrive to a proof. Denote by ${g = \frac{1}{n}(z_1+...+z_n)}$ . A quick computation using properties of the modulus gives:

$\displaystyle \sum_{k=1}^n |z-z_k|^2 = n|z-g|^2+\sum_{i=1}^n |z_k|^2$

Thus ${\sum_{k=1}^n |g-z_k|^2 = \sum_{i=1}^n |z_k|^2}$ . Of course, the classical inequality ${a^2+b^2 \geq 2ab}$ implies

$\displaystyle 2\sum_{k=1}^n |z_k|^2 = \sum_{k=1}^n |g-z_k|^2+\sum_{i=1}^n |z_k|^2\geq 2 \sum_{k=1}^n |z_k||g-z_k|.$

If the inequality in the statement of the problem holds, the above relation becomes an equality and ${|z_k|=|g-z_k|}$ for all ${k=1,...,n}$ . Therefore points ${z_k}$ belong to the mediatrix of the segment ${0g}$ . Therefore the centroid ${g}$ also belongs to this mediatrix and to ${0g}$ , which implies ${g=0}$ , as requested.

Conversely, if ${z_1+...+z_k = 0}$ consider the inequality

$\displaystyle |a||b| \geq \frac{1}{2}(\overline a b + a\overline b)$ to conclude.

Categories: Complex analysis, Geometry, Inequalities, Olympiad Tags: complex, Geometry, inequality, olympiad

Romanian Regional Olympiad 2024 – 10th grade

March 12, 2024 beni22sof Leave a comment

Problem 1. Let ${a,b \in \Bbb{R}}$ , ${a>1}$ , ${b>0}$ . Find the smallest real number ${\alpha}$ such that

$\displaystyle (a+b)^x \geq a^x+b, \forall x \geq \alpha.$

Problem 2. Consider ${ABC}$ a triangle inscribed in the circle ${\mathcal C}$ of center ${O}$ and radius ${1}$ . For any ${M \in \mathcal C\setminus \{A,B,C\}}$ denote by ${S(M) = OH_1^2+OH_2^2+OH_3^2}$ where ${H_1,H_2,H_3}$ are the orthocenters of the triangles ${MAB,MBC,MCA}$ , respectively.

a) Prove that if the triangle ${ABC}$ is equilateral then ${s(M)=6}$ for every ${M \in \mathcal C \setminus \{A,B,C\}}$ .

b) Show that if there exist three distinct points ${M_1,M_2,M_3 \in \mathcal C\setminus \{A,B,C\}}$ such that ${s(M_1)=s(M_2)=s(M_3)}$ , then the triangle ${ABC}$ is equilateral.

Problem 3. Let ${a,b,c}$ be three non-zero complex numbers with the same modulus for which ${A=a+b+c}$ and ${B=abc}$ are real numbers. Show that for every positive integer ${n}$ the number ${C_n = a^n+b^n+c^n}$ is real.

Problem 4. Let ${n \in \Bbb N^*}$ . Determine all functions ${f:\Bbb{R} \rightarrow \Bbb{R}}$ which verify

$\displaystyle f(x+y^{2n})=f(f(x))+y^{2n-1}f(y),$

for every ${x,y \in \Bbb{R}}$ and such that ${f(x)=0}$ has a unique solution.

Hints:

Problem 1: study the monotonicity of the function ${g(x)= (a+b)^x-a^x}$ . Then observe that the inequality is equivalent to ${g(x) \geq g(1)}$ .

Problem 2: Recall the identity ${OH^2 = 9R^2-AB^2-BC^2-CA^2}$ whenever ${H}$ is the orthocenter of ${ABC}$ with circumcenter ${O}$ . This can be proved using complex numbers and recalling that ${OH = 3OG}$ , where ${G}$ is the center of gravity. Therefore

$\displaystyle OH_1^2 = 9-MA^2-MB^2-AB^2$ and the analogue equalities. Summing we get

$\displaystyle s(M) = 27-2\sum MA^2-\sum AB^2.$ a) When ${ABC}$ is equilateral and inscribed in a circle of radius ${1}$ we have ${AB=BC=CA=\sqrt{3}}$ . Moreover, the identity In particular, one can prove the following:

$\displaystyle AM^2+BM^2+CM^2 = AG^2+BG^2+CG^2+3MG^2$ applied to the triangle equilateral triangle ${ABC}$ with centroid ${O}$ shows that

$\displaystyle MA^2+MB^2+MC^2 = AO^2+BO^2+CO^2+3MO^2=6.$ Thus

$\displaystyle s(M) = 27-12-9 = 6.$ b) Assume there exist three distinct points such that ${s(M_1)=s(M_2)=s(M_3)}$ . This implies that

$\displaystyle \sum M_1A^2 = \sum M_2A^2 = \sum M_3A^2.$ The relation above concerning the center of gravity of ${ABC}$ shows that ${M_1G=M_2G=M_3G}$ . Since the points are distinct, it follows that ${G}$ coincides with ${O}$ , the circumcenter of ${ABC}$ , therefore ${ABC}$ is equilateral.

Problem 3: Denote ${r>0}$ the common value of the modulus: ${|a|=|b|=|c|=r}$ . Then

$\displaystyle \overline{a+b+c} = r^2\left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) = r^2 \frac{ab+bc+ca}{abc}.$ Since ${r, a+b+c, abc \in \Bbb{R}}$ it follows that ${ab+bc+ca\in \Bbb{R}}$ . Then of course ${a^2+b^2+c^2 = (a+b+c)^2-2(ab+bc+ca) \in \Bbb{R}}$ . Finally, we know that ${a,b,c}$ are roots of

$\displaystyle (z-a)(z-b)(z-c)=0 \Longleftrightarrow z^3-(a+b+c)z^2+(ab+bc+ca)z-abc=0.$ Since the coefficients of this polynomial are real, an inductive argument shows that if ${C_n, C_{n+1}, C_{n+2}}$ are real then ${C_{n+3}}$ is real, finishing the proof.

Problem 4. Take ${y=0}$ and get ${f(x) = f(f(x))}$ . Thus, ${f}$ is the identity mapping on its image!! Take ${y\mapsto -y}$ and observe that ${y^{2n-1}f(y) = -y^{2n-1}f(-y)}$ . Therefore ${f(-y)=-f(y)}$ for any ${y \neq 0}$ . Since the equation ${f(x)=0}$ has a unique solution, it follows that ${f(0)=0}$ and ${f(x) \neq 0}$ for ${x \neq 0}$ . Take ${x=0}$ and get ${f(y^{2n}) = y^{2n-1}f(y)}$ . Therefore

$\displaystyle f(x+y^{2n})=f(x)+f(y^{2n})$ for any ${x,y}$ . Since ${y^{2n}}$ takes all positive values in ${\Bbb{R}}$ it follows that

$\displaystyle f(x+y) = f(x)+f(y)$ for every ${x\in \Bbb{R}}$ , ${y \geq 0}$ . Coupled with ${f(-y)=-f(y)}$ this implies that

$\displaystyle f(x+y)= f(x)+f(y).$ for all ${x,y}$ . If ${f(x_1)=f(x_2)}$ it follows that ${f(x_1-x_2)=0}$ therefore ${x_1=x_2}$ . From ${f(f(x))=f(x)}$ we obtain ${f(x)=x}$ for all reals ${x}$ . It should be noted that this function obviously verifies the functional equation!

Categories: Algebra, Geometry, Olympiad Tags: Algebra, complex, Geometry, olympiad

IMC 2013 Problem 6

August 11, 2013 beni22sof Leave a comment

Problem 6. Let ${z}$ be a complex number with ${|z+1|>2}$ . Prove that ${|z^3+1|>1}$ .

Solution: Write ${z}$ in its trigonometric form:

$\displaystyle z=r(\cos \theta + i \sin \theta).$

Then the condition ${|z+1|>2}$ translates to

$\displaystyle r^2+2r\cos \theta +1 >4$

which is equivalent to

$\displaystyle r^2+2r\cos \theta -3 >0.$

Since ${r}$ must be positive, we have as a consequence

$\displaystyle r > \sqrt{\cos^2 \theta+3}-\cos \theta.$

The condition ${|z^3+1|>1}$ is equivalent to

$\displaystyle r^6+2r^3 \cos (3\theta) >0$

which translates to

$\displaystyle r^3+2 \cos (3\theta)>0.$

Thus it is enough to prove that

$\displaystyle (\sqrt{\cos^2 \theta+3}-\cos \theta)^3 \geq -2 \cos (3\theta)=-2\cos \theta (4 \cos^2 \theta -3).$

Note that the result we need to prove is not trivial if and only if ${\theta \in (\pi/6,\pi/2)}$ and on that interval ${\cos \theta}$ is positive.

Denote ${t=\cos \theta}$ then ${t \in (0,\sqrt{3}/2)}$ and we want to prove that

$\displaystyle (\sqrt{t^2+3}-t)^3 +2t(4t^2-3)\geq 0.$

If we denote ${f(t)=(\sqrt{t^2+3}-t)^3 +2t(4t^2-3)}$ then ${f(\sqrt{3/8})=0=f'(\sqrt{3/8})=0}$ .

If we evaluate then we obtain

$\displaystyle f(t)=(a^2+3)^{3/2}+3a^2\sqrt{a^2+3}+4a^3-15a$

which is a sum of two convex functions (first two and last two terms) among which one is strictly convex, and therefore is strictly convex (on ${[0,1]}$ ).

This means that ${\sqrt{3/8}}$ is the global minimum of ${f}$ on ${[0,1]}$ and therefore ${f}$ is positive on that interval which finishes the proof.

Categories: Complex analysis, Olympiad Tags: complex, IMC

IMC 2013 Problem 5

August 8, 2013 beni22sof Leave a comment

Problem 5. Does there exist a sequence ${(a_n)}$ of complex numbers such that for every positive integer ${p}$ we have that ${\sum_{n=1}^\infty a_n^p}$ converges if and only if ${p}$ is not a prime?

Categories: Analysis, Complex analysis, Olympiad Tags: Analysis, complex, complex numbers, IMC

Geometric mean for n complex numbers

January 29, 2011 beni22sof Leave a comment

Let $D$ be a closed disc in the complex plane. Prove that for all positive integers $n$ , and for all complex numbers $z_1,z_2,\ldots,z_n\in D$ there exists a $z\in D$ such that $z^n = z_1\cdot z_2\cdots z_n$ .

Romanian TST 2004

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Categories: Complex analysis, Olympiad, Problem Solving Tags: complex, polynomial

Every expansion has a 0

October 13, 2009 beni22sof Leave a comment

Suppose $f$ is entire such that for each $x_0 \in \mathbb{C}$ at least one of the coefficients of the expansion $f=\sum\limits_{n=0}^\infty c_n (z-z_0)^n$ is equal to 0. Prove that $f$ is a polynomial.
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Categories: Complex analysis Tags: complex, entire, series, taylor

Power series & number theory

October 13, 2009 beni22sof Leave a comment

Let $F(z)=\sum\limits_{n=1}^\infty d(n) z^n$ where $d(n)$ denotes the number of divisors of $n$ . Calculate the radius of convergence of this series and prove that $F(z)=\sum\limits_{n=1}^\infty \frac{z^n}{1-z^n}$ . Furthermore, prove that $F(r)\geq \frac{1}{1-r}\log \frac{1}{1-r}$ for $r \in (0,1)$ .

Categories: Complex analysis, Number theory Tags: complex, divisor, series

Amazing property of entire functions

October 12, 2009 beni22sof Leave a comment

Prove that if $p,q,r$ are non-constant, non-vanishing entire functions with $p+q+r=0$ then there exists an entire function $h$ such that $p,q,r$ are constant multiples of $h$ .
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Categories: Complex analysis Tags: complex, entire, multiples

Blaschke factors

October 12, 2009 beni22sof Leave a comment

Take $w,z \in \mathbb{C}$ such that $\bar{z}w \neq 1$ . Prove that $\left| \frac{w-z}{1-\bar{w}z} \right|<1$ if $|z|<1$ and $|w|<1$ and also $\left| \frac{w-z}{1-\bar{w}z} \right|=1$ if $|z|=1$ or $|w|=1$ .

Prove that for any $w$ fixed in the unit disk $\mathbb{D}$ , the mapping $F_w(z)=\frac{w-z}{1-\bar{w}z}$ has the following properties:
1) $F_w$ maps $\mathbb{D}$ to $\mathbb{D}$ and is holomorphic.
2) $F_w$ interchanges $0$ and $w$ .
3) $|z|=1 \Rightarrow |F_w(z)|=1$ .
4) $F_w$ is an involution on $\mathbb{D}$ and thus is bijective.
5) Construct a bijective, holomorphic function, which maps the unit disk to itself, takes the unit circle into itself and interchanges two given complex numbers $z,w$ with $|z|<1,|w|<1$ .