Gheorghe ANDREI - COMPLEX NUMBERS (part II) - Some Problems -- 4 / Gheorghe ANDREI - NUMERE COMPLEXE (partea II) - Câteva Probleme -- 4 / Gheorghe ANDREI – LICZBY ZŁOŻONE (część II) – Niektóre problemy – 4

 I wrote about the book here.

                       We discuss some Identities with two or more complex variables. 

               I took a look at the classic works, of which I have the following:

                       @ EVGRAFOV M., BÉJANOV K., SIDOROV Y., FÉDORUK M, CHABOUNINE M.

RECUEIL DE PROBLÈMES SUR LA THÉORIE DES FONCTIONS ANALYTIQUES

MIR, Moscou, 1974

                     @@ Problems in the theory of functions of a complex variable , authors        L. Volkovysky, G. Lunts, I. Aramanovich (translated from the Russian by Victor Shiffer)  Mir ; Moscow; 1972 (I only have it in paper format)

                   @@@ ВОЛКОВЫСКИЙ Л. И., ЛУНЦ Г. Л., АРАМАНОВИЧ И. Г.

Сборник задач по теории функций комплексного переменного

ФИЗМАТЛИТ, Москва, 2004

                @@@@ VOLKOVYSKY I., LUNTS G., ARAMANOVICH I.

Problems in the Theory of Functions of a Complex Variable

MIR, Moscow, 1977 

              @@@@@ VOLKOVYSKII L[ev] I[zrailevich], LUNTS, G[rigorii] L[‘vovich],

             ARAMANOVICH I[saak] G[enrikhovich]

A Collection of Problems on Complex Analysis

DOVER PUBLICATIONS, INC. New York, 1991

    

.....................................................................................................................................

          From  @   we take problem 1.68 from page 19 and 1.59 from page 18 :

         1.68.  Show that for all complex values ​​of  $z$  and  $\zeta$  the following equalities are valid :

   1. $|z+\zeta|^2+|z-\zeta|^2=2|z|^2+2|\zeta|^2$

   2. $|z\bar \zeta+1|^2+|z-\zeta|^2=(1+|z|^2)(1+|\zeta|^2)$

   3. $|z\bar \zeta-1|^2-|z-\zeta|^2=(|z|^2-1)(|\zeta|^2-1)$

 

           1.59.  Show that for no matter what complex number  $z$,  the formula below is valid :

$|\sqrt{z^2-1}+z|+|\sqrt{z^2-1}-z|=|z-1|+|z+1|.$

            From  @@  we take problems 9, 10, 13  from page 12 :

             9.  Prove the identity :

$|z_1+z_2|^2+|z_1-z_2|^2=2(|z_1|^2+|z_2|^2.$

           10.  Prove the identity :

$|1-\bar z_1z_2|^2-|z_1-z_2|^2=(1-|z_1|^2)(1-|z_2|^2).$

           13.  Prove the foloowing identities :

  $$1)\;\;(n-2)\sum_{k=1}^n|a_k|^2+\bigg |\sum_{k=1}^na_k\bigg|^2=\sum_{1\leqslant k<s\leqslant n}|a_k+a_s|^2;$$

$$2)\;\;n\sum_{k=1}^n|a_k|^2-\bigg | \sum_{k=1}^na_k \bigg |^2=\sum_{1\leqslant k<s\leqslant n}|a_k-a_s|^2.$$

We will list from the basic work in the title the issues more or less related to this topic.

          Problem 35* (page 25)  Show that for any  $z\in \mathbb{C}$  the relation holds :

$|\sqrt{z^2-1}+z|+|\sqrt{z^2-1}-z|=|z-1|+|z+1|.$

          Problem 53 (page 27)  Show that :

$2|\sqrt{z^2-1}+z|=|z+1|+|z-1|+\sqrt{(|z+1|^2+|z-1|^2)-4},\;\;\;\forall z\in \mathbb{C}.$

          Problem 55 (page 27)   If  $z\in \mathbb{C}$  then

$|\sqrt{z^2+2z}+z+1|+|\sqrt{z^2+2z}-(z-1)|+|z-2|=$

$=|\sqrt{z^2-2z}+z-1|+|\sqrt{z^2-2z}-(z-1)|+|z+2|.$

          Problem  1 (page 28-29)   To check the equalities :

a)  $|z_1+z_2|^2+|z_1-z_2|^2=2(|z_1|^2+|z_2|^2);$

b)  $|z_1|^2+|z_2|^2-z_1 \bar z_2-\bar z_1 z_2=|z_1-z_2|^2;$

c)  $|z_1\bar z_2+1|^2+|z_1-z_2|^2=(|z_1|^2+1)(|z_2|^2+1);$

d)  $|z_1 \bar z_2-1|^2-|z_1-z_2|^2=(|z_1|^2-1)(|z_2|^2-1);$

e)  $|z_1+2z_2|^2-2|z_1+z_2|^2=2|z_2|^2-|z_1|^2.$

          Problem  2 (page 29)  If  $z_1,\;z_2\in \mathbb{C}$,  then :

a)  $|z_1 \bar z_2+1|^2+|z_2 \bar z_1+1|^2+2|z_1-z_2|^2=2(1+|z_1|^2)(1+|z_2|^2);$

b)  $z_1 \bar z_2-1|^2+|z_2 \bar z_1-1|^2-2|z_1-z_2|^2=2(1-|z_1|^2)(1-|z_2|^2);$

c)  $|1-\bar z_1z_2|^2-|z_1-z_2|^2=(1+|z_1z_2|)^2-(|z_1|+|z_2|)^2.$

          Problem  4 (page 29)  BERGSTROM identity : 

 for  $u,\;v \in \mathbb{C}\;\;and \;\;a,\;b\in \mathbb{R}^*,\;a+b \neq 0$

$\frac{|u|^2}{a}+\frac{|v|^2}{b}-\frac{|u+v|^2}{a+b}=\frac{|bu-av|^2}{ab(a+b)}.$

ANSWER CiP / Solutions CiP

  

                    @1.68 1) or @@9  or  #1a) (p. 28)        $|z_1+z_2|^2+|z_1-z_2|^2=$

$=(z_1+z_2)\cdot \overline{(z_1+z_2)}+(z_1-z_2)\cdot \overline{(z_1-z_2)}=(z_1+z_2)(\bar z_1+\bar z_2)+(z_1-z_2)(\bar z_1-\bar z_2)=$

$=(z_1 \bar z_1+z_2 \bar z_1+z_2\bar z_1+z_2 \bar z_2)+(z_1 \bar z_1-z_1\bar z_2-z_2 \bar z_1+z_2 \bar z_2)=$

$=2z_1 \bar z_1+2z_2 \bar z_2=2(|z_1|^2+|z_2|^2)$

(to be continue)