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andrei_g
Pe: 2011-10-29T13:05:14+03:00 In: In:

Forma trigonometrica a unui numar complex

1. Sa se arate ca modulul unui numar complex

    \[ z = \frac{{1 + \lambda i}}{{1 - \lambda i}},\;\lambda \in R \]

,este egal cu 1. Reciproc, sa se arate ca orice numar complex de modul 1 si diferit de -1 poate fi scris in mod unic sub forma:

    \[ \frac{{1 + \lambda i}}{{1 - \lambda i}},\;\lambda \in R\]

.

2.Sa se calculeze modulul si argumentul numarului complex:

    \[ \begin{array}{l} a).\quad z_1 = \frac{1}{{1 + i \cdot tg\;\alpha }}\quad ,\alpha \in R - \left\{ {\frac{{(2k + 1)\pi }}{2}|k \in Z} \right\} \\ \\ b).\quad z_2 = a\frac{{1 + ib}}{{1 - ib}}\;,a,b \in R. \\ \end{array} \]

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1 raspuns

  1. PhantomR
    expert (VI)

    1.
    "\Rightarrow"
    z=\frac{1+\lambda i}{1- \lambda i }=\frac{(1+ \lambda i)^2}{(1-\lambda i)(1+\lambda i)}=\frac{1-\lambda ^2}{1+\lambda ^2}+i \frac{2\lambda}{1+\lambda^2} \Rightarrow |z|=\sqrt{\frac{1^2-2\lambda ^2+\lambda ^4+4\lambda ^2}{(1+\lambda)^2}}=\sqrt{\frac{(1+\lambda ^2)^2}{(1+\lambda ^2)^2}}=\sqrt{1}=1.

    "\Leftarrow"
    EXISTENTA: Fie z=a+bi cu a,b\in\mathbb{R}. Atunci |z|=1 \Leftrightarrow \sqrt{a^2+b^2}=1 \Leftrightarrow a^2+b^2=1. Putem alege a=\frac{1-\lambda^2}{1+\lambda^2} si b=\frac{2 \lambda}{1+\lambda^2} si conform calcului de la implicatia directa rezulta cerinta.

    UNICITATEA: Presupunem prin absurd ca ar exista \alpha\neq \lambda cu \alpha,\lambda\in\mathbb{R} astfel incat astfel incat z=\frac{1+\lambda i}{1-\lambda i}=\frac{1+\alpha i}{1- \alpha i} \Rightarrow (1+\lambda i)(1-\alpha i)=(1-\lambda i)(1+\alpha i) \Rightarrow 1-\alpha i + \lambda i + \lambda \alpha = 1+ \alpha i - \lambda i + \alpha \lambda \Rightarrow 2\alpha i = 2 \lambda i \Rightarrow\alpha=\lambda, contradictie. Asadar scrierea este unica.

    2.
    a)
    z_1=\frac{1}{1+itg\alpha}=\frac{1}{1+i\frac{sin\alpha}{cos\alpha}}=\frac{cos\alpha}{cos\alpha+isin\alpha}=\frac{cos\alpha(cos\alpha-isin\alpha)}{cos^2\alpha+\sin^2alpha}=cos^2\alpha+isin\alpha cos\alpha. Cum \alpha\in\mathbb{R}-\{\frac{(2k+1)\pi}{2}|k\in\mathbb{Z}\} avem cos\alpha \neq 0, deci argz_1=arctg\frac{sin\alpha cos\alpha}{cos^2 \alpha}=arctg \frac{sin\alpha}{cos\alpha}=arctgtg\alpha=\alpha.

    |z_1|=\sqrt{cos^4\alpha+sin^2\alpha cos^2\alpha}=\sqrt{cos^2\alpha(cos^2\alpha + sin^2 \alpha)}=\sqrt{cos^2\alpha}=|cos\alpha|.

    b) z_2=a\frac{1+ib}{1-ib} \Rightarrow |z_2|=|a\frac{1+ib}{1-ib}|=|a|\cdot|\frac{1+ib}{1-ib}|=|a|. Am folosit rezultatul de la punctul 1. Demonstratia se gaseste mai sus.

    Facand calculele ca la punctul 1. obtinem: z_2=a\frac{1-b^2}{1+b^2}+ia\frac{2b}{1+b^2}. Daca a=0, atunci z_2=0 si nu are argument, caci nu exista x\in\mathbb{R} astfel incat \sin x = \cos x = 0. Daca 2b=0 \Leftrightarrow b=\frac{0}{2} \Leftrightarrow b=0 atunci argz_2=arctg 0= 0. Daca a\neq 0 si 1-b^2=0 \Leftrightarrow 1=b^2 \Leftrightarrow b=\pm1 atunci z_2 este pur imaginar si argumentul sau este: argz_2=\begin{array} -\frac{\pi}{2}, Im z_2<0 \\ \frac{\pi}{2}, Im z_2>0 \end{array}.

    Excluzand cazurile de mai sus avem: arg z_2 = arctg \frac{a\frac{2b}{1+b^2}}{a\frac{1-b^2}{1+b^2}}=arctg \frac{2b}{1-b^2}.

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