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 

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 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>