GAZETA MATEMATICĂ Seria B N0 4/2023

 Faceti click pe imagine pentru descarcare.

        Vezi pag. 195 rezolvarea problemei E:16386.

SUPLIMENTUL cu EXERCIȚII al GMB N0 4/2023

 Faceti click pe imagine pentru descarcare

Vezi ERATA

The Equation of a Straight Line in different Analytical Geometries

          We denote with $\mathbb{R}$ the set of real numbers. Sometimes it means the field of reals , with the usual operations of addition $"+"$ and multiplication $"\cdot"\;$; we note its elements with small Greek letters $\;\alpha,\;\beta,\;\gamma \dots$

          With $\mathbb{C}$ we denote the set of complex numbers $z=\alpha +\imath \beta$; its elements are usually denoted by small Latin letters $\;a,\;b,\;c\dots$ The complex number $\alpha -\imath \beta$, written $\bar{z}$ is the complex conjugate of   $z$. We denote $\alpha=\Re(z)$ and $\beta=\Im(z)$ respectively the real part and the imaginary part of $z$.

          By $\mathbb{A}^2$ we denote the real 2-dimensional affine space or plane ; in it we consider a Cartesian coordinate system (affine frame) $\mathscr{R}=(O;\;\mathbf{e}_1,\mathbf{e}_2)$, where $O$ is a fixed point and $\;\mathbf{e}_1,\mathbf{e}_2$ a base of (real) vectorial space $\mathbb{R}^2$. Endowed with a dot product he becomes the Euclidean plane $\mathbb{E}^2$; in this particular case we prefer the base vectors to be orthonormal.

               In the Proposition below, $\mathbb{C}$ is considered as the 2-dimensional real affine space $\mathbb{R}\times \mathbb{R}$ with the affine frame $\mathscr{S}=(O;\;1,\imath)$. The point $O$ is the same as the one in $\mathscr{R}$ above.

           %1.     PROPOSITION  (i)  The straight line in $\mathbb{E}^2$ that corresponds in the

                               frame $\mathscr{R}$ to the equation

$$\alpha \cdot x+\beta \cdot y+\gamma=0\;\;\;(\alpha,\beta)\neq (0,0) \tag{1}$$

                                has in the frame $\mathscr{S}$ of $\mathbb{C}$ the equation

$$\frac{\alpha+\imath \beta}{2}\cdot z+\frac{\alpha-\imath \beta}{2} \cdot \bar{z}+\gamma=0.\tag{2}$$

                                  (ii) Converselly, to any equation in the frame $\mathscr{S}$ of the form 

$$a\cdot z+b \cdot \bar{z}+c=0,\;\;0\neq b=\bar{a},\;c=\bar{c} \tag{3}$$

                                corresponds a straight line which in the frame $\mathscr{R}$ has the equation

$$(2\cdot \Re\;a)\cdot x+(2\cdot \Im\;a)\cdot y +c=0.\tag{4}$$

          The most general case of an equation representing a straight line is given in the following theorem. ($|a|=\sqrt{\alpha^2+\beta^2}$ means the modulus of the complex number $a=\alpha+\imath \cdot \beta$.)

               %2. THEOREM  The equation

$$a\cdot z+b \cdot \bar {z}+c=0,\;\;\;(a,b)\neq (0,0) \tag{5}$$

           represents a straight line if and only if

 $$|a|=|b|\;\;and\;\;\bar {a} \cdot c=b \cdot \bar{c}. \tag{6}$$

               %3 CORROLARY  (i) If  $c=0$ then equation (5) represents a line iff

 $|b|=|a|. \tag{6i}$

                                                  (ii) If $c \neq 0$ then equation (5) represents a line iff

 $b \cdot \bar{c}=\bar{a} \cdot c.\tag{6ii}$

                        Indeed, if $c=0$, then the second condition in (6) is automatically satisfied. If $c\neq 0$, then $|b|=\left |\frac{\bar{a}\cdot c}{\bar{c}} \right |=\frac{|\bar{a}|\cdot |c|}{|\bar{c}|}=|\bar{a}|=|a|$, and the conditions (6) are both fulfilled.

$\square$

      Some authors prefer to write the equation of a line in the form

$b \cdot z-\bar{b}\cdot \bar{z}+c=0$, $c-$purely imaginary. $\tag{7}$

     The equation (3) is called the self-adjoint form. Since we can multiply equation (5) by an arbitrary non-zero complex number, any of the forms is equally justified when $c \neq 0$.

     Indeed, given a straight line (6) with $c\neq 0$

$$a\cdot  z+b\cdot \bar{z}+c=0$$

$$\underset{(6ii)}{\Rightarrow}\;\;a \cdot z+\frac{\bar{a}c}{\bar{c}} \cdot \bar{z}+c=0\;\;\Rightarrow$$

$$\Rightarrow\;\;a\bar{c}\cdot z+\bar{a}c \cdot \bar{z}+c\bar{c}=0;$$

because $c\bar{c}=|c|^2$ is real number, this is the self-adjoint form. Further, multiplying the last equation by $\imath$

$$\imath a\bar{c} \cdot z+\imath \bar{a}c \cdot \bar{z}+\imath |c|^2=0$$

and writing $\;b:=\imath a\bar{c}$, because $\imath \bar{a}c=(-\bar {\imath})\overline{(\bar{a})}\bar{c}=-\overline{\imath \bar{a}c}$, we get the form (7).

Let's note that, also in the case $c=0$, an equation $az+b\bar{z}=0$ in which the numbers $a$ and $b$ seem arbitrary (satisfying the condition (6i) $|a|=|b|$), can be brought to the self-adjoint form. If we multiply this equation by $\bar{a}+\bar{b}$, we get the equation 

$$a_1 \cdot z +b_1 \cdot \bar{z}=0 \tag{8}$$

in which $a_1=a(\bar{a}+\bar{b})$ and $b_1=b(\bar{a}+\bar{b})$. But the condition (6i) is also expressed as $a \cdot \bar{a}=b \cdot \bar{b}$ and then we have

$\bar{a_1}=\overline {a\bar{a}+a\bar{b}}=\bar{a}a+\bar{a}b=\bar{b}b+\bar{a}b=b(\bar{b}+\bar{a})=b_1.$

     Example: The equation of the straight line $z+\imath \cdot \bar{z}=0$, after multiplying by $1-\imath$, turns into $(1-\imath)z+(\imath +1)\bar{z}=0$, obviously a self-adjoint form (3).

          So (3) is the general form of the equation of a straight line.

{Last edit: 7/29/2023}

Text_ciorna :

$\mathscr{R}$

$\mathbf{e}$

and the FFF conditions are both fulfilled.

End_ciorna

एक पत्रिका जो दुनिया भर में घूमती है // A magazine that travels the world

           I mentioned this magazine before in the post of Wednesday, January 25, 2023.

           I received an email in the last days from a professor from Calcutta (India), interested in some Romanian magazines. He also asked me some details about "Revista Matematica din Timisoara". Unfortunately, I haven't scanned many numbers yet, but I hope that in the future I will be able to post the entire collection I have.