Gheorghe ANDREI - COMPLEX NUMBERS (part II) - Some Problems -- 3 / Gheorghe ANDREI - NUMERE COMPLEXE (partea II) - Câteva Probleme -- 3 / Gheorghe ANDREI - KOMPLEXNÉ ČÍSLA (časť II) - Niektoré problémy -- 3

 I wrote about the book here.

               We discuss problems no. ##3, 4, page 21 (solved on page 177) and #38, page 25 (solved on page 185).

                Problem #3  "Let  $\varepsilon$  be a complex cube root of unity. Prove that

$|z-1|^2+|z-\varepsilon|^2+|z_\varepsilon^2|^2=3(|z|^2+1)"$

                Problem #4   "Let  $\varepsilon$  be a complex cube root of unity. Prove that

$|z+u|^2+|z+\varepsilon u|^2+|z+\varepsilon^2 u|^2=3(|z|^2+|u|^2)$

                         whatever  $z,\;u\in\mathbb{C}$  are."

                                  Problem #38  "Prove that :

                                           a)   $(z-1)^2+(z-\varepsilon)^2+(z-\bar\varepsilon)^2=3z^2$

                where  $\varepsilon=\cos\frac{2\pi}{3}+\imath \sin\frac{2\pi}{3};$

                                          $\textbf{b)}\;\;\;(z-1)^2+(z-\varepsilon)^2+(z-\varepsilon ^2)^2+\dots+(z-\varepsilon ^{n-1})^2=nz^2$

where  $\varepsilon=\cos \frac{2\pi}{n}+\imath \sin \frac{2\pi}{n};$

 $$\textbf{c)}\;\;\; \sum_{k=0}^{n-1}|z-\varepsilon ^k|^2=n(|z|^2+1|)."$$

(to be continue)

Gheorghe ANDREI - COMPLEX NUMBERS (part II) - Some Problems -- 2 / Gheorghe ANDREI - NUMERE COMPLEXE (partea II) - Câteva Probleme -- 2 // Gheorghe ANDREI - اعداد مختلط (قسمت دوم) - برخی از مسائل - 2

 You can find the book in my Electronic Library. The author, Gh. ANDREI gets lost among the crowd of Titu ANDREESCU...(current version today ; you can still access the library by scanning the QR code from the beginning)

                                                                                       < begin preview>

ANDREESCU Titu, DOSPINESCU Gabriel, MUSHKAROV Oleg

NUMBER THEORY: CONCEPTS AND PROBLEMS

XYZ Press, Plano_TX, 2017

ANDREI Gheorghe

NUMERE COMPLEXE : partea a II-a

Ed. GIL, Zalău, 2004/2005-ebook(cadou de Craciun 25 Dec 2025 de la fiul meu

                                                                                                                     CIOBANU Victor)

ANDREIAN-CAZACU Cabiria, DELEANU Aristide, JURCHESCU Martin

TOPOLOGIE. CATEGORII. SUPRAFEȚE RIEMANNIENE

Ed. ACADEMIEI [R.S.R.], București, 1966 

< end preview>

 This post is number 2, because we believe that the post here would be number 1.

 

            Problem #2 (page 21, solved on page 177 ; in translation)

"Let  $z\in \mathbb{C}\setminus \{\pm 1\}$, with the property  $\imath \frac{z-1}{z+1}\in\mathbb{R}$. Determine  $|z|$."

ANSWER CiP

$$|z|=1$$

                    Solution CiP

                Let  $\mathbb{R}\ni a=\imath \cdot \frac{z-1}{z+1}.$  We have (because $\imath ^2=-1\Rightarrow \frac{1}{\imath}=-\imath$)

$\frac{z-1}{z+1}=\frac{a}{\imath}=-a\imath$

so

 $z=\frac{1-a\imath}{1+a\imath} \tag{1}$

hence, according to Property #20, page 13,

$|z|=\frac{|1-a\imath|}{|1+a\imath|}=\frac{\sqrt{1+a^2}}{\sqrt{1+a^2}}=1$

$\blacksquare$

                 REMARKS CiP

               $1^r$.  Continuing in (1), we have  $z=\frac{(1-a\imath)^2}{(1+a\imath)(1-a\imath)}$  so

$z= \frac{1-a^2}{1+a^2}-\frac{2a\imath}{1+a^2}\;\overset{b=-a}{=}\;\frac{1-b^2}{1+b^2}+\frac{2b\imath}{1+b^2}\tag{2}$

               $2^r$.  The reciprocal statement is also true :

$|z|=1\Rightarrow (\exists)a\in\mathbb{R}\;\;s.t.\;\;z=\frac{1-a^2}{1+a^2}-\frac{2a}{1+a^2}\imath\Leftrightarrow (\exists)b\in\mathbb{R}\;\;s.t.\;\;z=\frac{1-b^2}{1+b^2}+\frac{2b}{1+b^2}\imath \tag{3}$

Indeed, according to page 16, Trigonometric form of complex numbers, #1, we have

$z=\cos \theta+\imath \sin \theta$,  and  $\cos \theta=\frac{1-b^2}{1+b^2},\;\;\sin \theta =\frac{2b}{1+b^2},\;\;\;b=\tan \theta /2$

hence (3).

$\square$<end REM>

PROBLEM E : 12 955 ABOUT the LACK of PERFECT SQUARES

 In GMB 5/2005 magazine, page 230 :

          E:12955  Prove that for any  $x\in\mathbb{N}$, the number  $x^2+13x+40$  

                            is not a perfect square.

([Authors:] Ioana CRĂCIUN și Gh. CRĂCIUN, Plopeni)

To avoid complicated Diophantine equation techniques, we looked for the interleaving of the number  $x^2+13x+40$  between two consecutive squares.

ANSWER CiP

$(x+6)^2<x^2+13x+40<(x+7)^2 \tag{1}$

                           Solution CiP

$0<x+4<x+13\;\;\;\;|+(x^2+12x+36)$

$\Rightarrow x^2+12x+36<x^2+13x+40<x^2+14x+49\Rightarrow (x+6)^2<x^2+13x+40<(x+7)^2$

hence (1).

$\blacksquare$

PROBLEM C : 2870 from PROBLEMS for the ANNUAL COMPETITION of "GAZETA MATEMATICĂ"

          In GMB 5/2005 magazine, pages 234 (Romanian version) and 236 (English version).

          The problem says : 

     C:2870.  Find the minimum value of :  $E(x)=\frac{x^2-2x-1}{x^2-2x+3}$,  for  $x\in\mathbb{R}$.

[Author :] Vasile PREDAN, Curtea de Argeș

ANSWER CiP

$-1=E(1)\leqslant E(x)<1$

                     Solution CiP  We will imitate the solution from the Post here.

           The number  $\lambda$  is a value of  $E(x)\;\;\Leftrightarrow$

$\Leftrightarrow (\exists)x\in\mathbb{R}\;\;\mathbf{s.t.}\;\;E(x)=\lambda$

$\Leftrightarrow (\exists)x\in\mathbb{R}\;\;\mathbf{s.t.}\;\;x^2-2x-1=\lambda \cdot (x^2-2x+3)$

$\Leftrightarrow (\exists)x\in\mathbb{R}\;\;\mathbf{s.t.}\;\;(1-\lambda)x^2+2(\lambda-1)x-(3\lambda+1)=0 \tag{1}$

$\Leftrightarrow$  the quadratic equation (1) has solution(s).

[For  $\lambda=1$  equation (1) isn't quadratic both then in hasn't solution.].

     Using the half discriminant ($\Delta'=(b')^2-ac$) formula, the roots of equation (1) are

$x_{1,2}=\frac{1-\lambda\pm\sqrt{2-2\lambda^2}}{1-\lambda} \tag{2}$

The requirement that the roots be real numbers, $|\Delta'\geqslant 0$  implies  $2-2\lambda^2\geqslant 0\;\Leftrightarrow$

$\Leftrightarrow\;\lambda^2\leqslant 1\;\Leftrightarrow\;-1\leqslant \lambda<1.$ From (2) we obtain the expressions of the roots as follows  $x_{1,2}=1\pm \sqrt{\frac{2(1+\lambda)}{1-\lambda}}.$

$\blacksquare$

       Writing differently,  $E(x)=1-\frac{4}{(x-1)^2+2}\;\;\Rightarrow\;\; -1=1-\frac{4}{0+2}\leqslant 1-\frac{4}{(x-1)^2+2}<1$,  with the same answer.

A RATIONAL VALUE of COSINE // مقدار گویای کسینوس

               In the magazine GAZETA MATEMATICĂ No 4, 1978, on covers 3-4 the following Problem appears (as a solution to the one proposed in GMB 11/1977, page 455 Problem#3) :

          Problem 3 page 455 / No. 11/1977  : Prove that if   $\alpha \in \mathbb{R}$  and 

      $\cos(\alpha \pi)=\frac{1}{3}$  then  $\alpha$  is irrational. (The angle  $\alpha \pi$ is considered in radians)

(We will return to this issue at the end)

               In the past, in my youth, I was more diligent. In the magazine GMB 10/2014 (so 35 years after the one mentioned in the preamble), in the -recently established- Column PROBLEMS for NATIONAL EXAMS, in the 12th grade, page 478, we see problems 27 and 28. (We made a system of cards with them, by grade, etc.; that's why I said "diligence")

This statement

Solving

As you can see, in Problem 2c), the question arises whether at the value  $q\pi$  for which  $\cos q\pi=\frac{3}{5}$, we have  $q\in\mathbb{Q}$  or not. When resolving this point, I remembered the problem in the Preamble. The answer is NO.

So, $\cos q\pi=\frac{3}{5}\;\Rightarrow\; q\notin \mathbb{Q}$.

         We assume the opposite (see line (1) in the third picture) : 

$\exists q\in \mathbb{Q}\;\;s.t.\;\;\cos q\pi=\frac{3}{5}\tag{1}$

Let  $q=\frac{m}{n},\;m\in\mathbb{Z},\;n\in \mathbb{N}$, with their greatest common divisor  $(m,n)=1$. Then

 the set of values ​​of the string  $\{\cos kq\pi;\;q\in\mathbb{Z}\}$ is finite :                           (2)

$\pm\cos 0q\pi,\;\pm\cos q\pi,\;\pm\cos 2q\pi,\dots,\;\pm\cos (n-1)q\pi \tag{3}$

     Indeed, we will show that, whatever  $k\in\mathbb{Z},\;\cos kq\pi$  has one of the values  (3). Writing, with Euclid's division lemma,  $k=n\cdot t+l,\;\;t\in\mathbb{Z},\;l\in\{o,\;1,\;\dots,\;n-1\}$,  we have

$\cos kq\pi=\cos k(\frac{m}{n})\pi=\cos (nt+l)\frac{m}{n}\pi=\cos (mt+l\frac{m}{n})\pi=\cos (mt\pi+lq\pi)=\pm\cos lqt$

qed$\square$

       On the other hand, we will show that if  $k$ is a power of 2, then  $\cos k\pi$  takes on an infinity of values. More precisely, we show, by mathematical induction, the equality :

$\cos 2^kq\pi=\frac{a_k}{5^{2^k}},\;\;\;a_k\in\mathbb{Z},\;\;\;5\not{\mid}a_k \tag{4}$

For $k=0,\;\;\cos 2^0q\pi=\cos q\pi=\frac{3}{5}=\frac{3}{5^{2^0}}$, and we have  $a_0=3,\;5\not{\mid }a_0.$ Then  $\cos 2q\pi=2\cos ^2q\pi-1=2(\frac{3}{5})^2-1=-\frac{7}{25}=\frac{-7}{5^{2^1}}$, so  $a_1=-7\;and\;5\not{\mid}a_1$.

Assuming (4) true for a fixed  $k$, we have

$\cos 2^{k+1}q\pi=\cos 2(2^kq\pi)=2\cos^2(2^kq\pi)-1=2\cdot (\frac{a_k}{5^{2^k}})^2-1=\frac{2a_k^2-5^{2^k\cdot 2}}{5^{2^k\cdot 2}}=\frac{2a_k^2-5^{2^{k+1}}}{5^{2^{k+1}}}$

and taking  $a_{k+1}=2a_k^2-5^{2^{k+1}}$  we have  $5\not{\mid}a_{k+1}$  because  $5\not{\mid}a_k$. So (4) is thrue for  $k+1$. Moreover, the values ​​given by (4) are all distinct, because if we had  $\frac{a_k}{5^{2^k}}=\frac{a_l}{5^{2^l}},\;k<l$, it would result  $a_l=5^{2^l-2^k}\cdot a_k$  so  $5\mid a_l$ - false.

$\blacksquare$

               The same argument is used in the original problem in the Preamble, where, from  $\cos \alpha \pi=\frac{1}{3}$,  results :

$\cos 2^k\alpha \pi=\frac{r}{3^{2^k}},\;\;r\in\mathbb{Z},\;\;\;3\not{\mid}r$

Problem 2954 from REVISTA MATEMATICĂ a ELEVILOR din TIMIȘOARA

 You can see issue 1-1977 from which I took the issue (page 62), here.

A larger collection is here.

             2 954.  Determine the real numbers  $x$  for which the number

$\frac{x}{x^2-5x+7}$  is an integer.

(Matematika v Șkole)

ANSWER CiP

$x\in \{0,\;2,\;\frac{7}{3},\;3,\;\frac{7}{2}\}$

                    Solution CiP

          If  $\frac{x}{x^2-5x+7}=k\in\mathbb{Z}$  then  $x$  is the root of the quadratic equation

$kx^2-(5k+1)x+7k=0 \tag{1}$

The equation (1) has the discriminant  $\Delta_x=(5k+1)^2-28k^2=1+10k-3k^2$.

     But  $1+10k-3k^2=\frac{28}{3}-\left (\frac{25}{3}-10k+3k^2\right )=\frac{28}{3}-3\cdot \left(\frac{25}{9}-\frac{10}{3}k+k^2\right)=$

$=\frac{28}{3}-3\cdot \left (\frac{5}{3}-k\right)^2\leqslant \frac{28}{3}$,  so  

$\Delta_x\leqslant \frac{28}{3}\tag{2}$

     For the equation (1) to have rational roots, it must be  $\sqrt{\Delta_x}\in\mathbb{Q}$,  in fact  $\sqrt{\Delta_x}\in\mathbb{N}$. This happens if  (cf. (2))

$\sqrt{\Delta_x}\in\{0,\;1,\;2,\;3\} \tag{3}$

     Solving the equations  $1+10k-3k^2=0\;or\;1\;or\;2\;or\;3$  we obtain the integer values

$k\in\{0,\;2,\;3\}$

for which, we obtain from the equation (1) the values ​​of  $x$  in the answer.

$\blacksquare$

          Correction

          My answer is incomplete, as the commenter noted. 

          For  $x=3\pm\sqrt{2}$, the number  $\frac{x}{x^2-5x+7}$  takes the value  $1$, also an integer.

I wrongly assumed that  $x$  is a rational number. Let's analyze the sign of  $\Delta_x$.

          Setting the condition  $\Delta_x \geqslant 0\;\Rightarrow\; 1+10k-3k^2\geqslant 0$  which gives us the admissible values

$k\in\left ( \frac{5-\sqrt{28}}{3}\;,\;\frac{5+\sqrt{28}}{3}\right )\cap \mathbb{Z}=\{0,\;\color{Red}1,\;2,\;3\}$

Considering that the roots of the equation (1) are  $x_{1,2}=\frac{5k+1\pm\sqrt{\Delta_x}}{2k}$, for  $k=1$  we find  $x=3\pm\sqrt{2}$, so the correct answer is

$x\in \{0,\;3-\sqrt{2},\;2,\;\frac{7}{3},\;3,\;\frac{7}{2},\;3+\sqrt{2}\}$

$\color{Red}{!!!}\square \color{Red}{!!!}$

PROBLEM E : 6185 from the magazine GAZETA MATEMATICĂ

          From GMB 4/1978 page 164.

In translation :

                         "E : 6185*.  Knowing that the numbers  4830,  448230,  44482230

         are products of two consecutive natural numbers, find these consecutive

                      numbers. Generalization.

[Author : ] I. I. Mihailov, Ivanovo, U.R.S.S."

ANSWER CiP

$4\;830\;=\;69\;\times\;70$

$448\;230\;=\;669\;\times\;670$

$44\;482\;230\;=\;6\;669\;\times\;6\;670$

[CiP addition] $4\;444\;822\;230\;=\;66\;669\;\times\;66\;670$

                                  Generalization :

$\underset{n+1}{\underbrace{4\;\dots\;4}}\;8\;\underset{n}{\underbrace{2\;\dots\;2}}\;30\;=\;\underset{n+1}{\underbrace{6\;\dots\;6}}\;9\times \underset{n}{\underbrace{6\;\dots\;6}}\;70 \tag{G}$

                         Solution CiP

           If a number  $A$  is a product of consecutive numbers, i.e.  $A=k(k+1)$, then we have

$A=k(k+1) \Rightarrow k^2<A<(k+1)^2\Rightarrow k<\sqrt{A}<k+1\Rightarrow k=[\sqrt{A}]$

([X]=the integer part of the number X) We have :

$\sqrt{4830}=69,4...$  and it is verified by calculation that  $4830=69\times 70$

$\sqrt{448230}=669,4...$  and it is verified by calculation that  $448230=669\times 670$

$\sqrt{44482230}=6669,4...$  and it is verified by calculation that  $44482230=6669\times 6670$

The next example in the answer, verified by calculation, is my creation. I also formulated the generalization, which we will verify below.

          To carry out the calculations we need to replace  $1\;\dots\;1$  with the more precise expression

$\underset{n}{\underbrace{1\;\dots\;1}}=\frac{1}{9}\cdot (10^n-1) \tag{1}$

Indeed  $\underset{n}{\underbrace{1\;\dots\;1}}=\frac{1}{9}\cdot \underset{n}{\underbrace{9\;\dots\;9}}=\frac{1}{9}\cdot (1\underset{n}{\underbrace{0\;\dots\;0}}-1)=\frac{1}{9}\cdot(10^n-1).$

          Analyzing the position of the digits of the number  $4\dots482\dots230$, we obtain that the value of the number on the left of the relation  (G) is :

$\underset{n+1}{\underbrace{4\;\dots\;4}}\underset{n+3}{\underbrace{\;8\;\overset{n+2}{\overbrace{\underset{n}{\underbrace{2\;\dots\;2}}\;\underset{2}{\underbrace{30}}}}}}=\underset{n+1}{\underbrace{4\dots4}}\cdot 10^{n+3}+8\cdot 10^{n+2}+\underset{n}{\underbrace{2\dots2}}\cdot 100+30=$

$\overset{(1)}{=}4\cdot \frac{1}{9}\cdot (10^{n+1}-1)\cdot 10^{n+3}+8\cdot 10^{n+2}+2\cdot \frac{1}{9}\cdot(10^n-1)\cdot 100+30=$

$=\frac{4}{9}\cdot 10^{2n+4}-\frac{4}{9}\cdot 10^{n+3}+8\cdot 10^{n+2}+\frac{2}{9}\cdot 10^{n+2}-\frac{2}{9}\cdot 100+30=$

$=\frac{4}{9}\cdot 10^{2n+4}+\frac{34}{9}\cdot 10^{n+2}+\frac{70}{9} \tag{2}$

     On the right side of (G) we have :

$\underset{n+1}{\underbrace{6\dots6}}9\times \underset{n}{\underbrace{6\dots6}}70=\left [\frac{6}{9}\cdot (10^{n+1}-1)\cdot 10+9\right ]\times \left[\frac{6}{9}\cdot (10^n-1)\cdot 100+70\right ]=$

$=\left [\frac{2}{3}\cdot 10^{n+2}+\frac{7}{3}\right]\cdot \left [\frac{2}{3}\cdot 10^{n+2}+\frac{10}{3}\right ]=\frac{4}{9}\cdot 10^{2n+4}+\frac{2}{3}\cdot 10^{n+2}\cdot \left (\frac{7}{3}+\frac{10}{3}\right )+\frac{70}{9}=$

$=\frac{4}{9}\cdot 10^{2n+4}+\frac{34}{9}\cdot 10^{n+2}+\frac{70}{9} \tag{3}$

The results in (1) and (2) prove the equality in (G).

$\blacksquare$