×   HOME JAVA NETPLOT OCTAVE Traži ...
  matematika1
Dijeljenje kompleksnih brojeva     OSNOVE MATEMATIKE     Kompleksna ravnina


Jednadžbe u skupu kompleksnih brojeva

Riješite jednadžbe:
a)
$ \displaystyle \frac{(3+2i)(1+i)+2i}{(2-i)(1+i)-3}=\frac{7-i}{-4}\, z^4$ ,

b)
$ \displaystyle \left[\frac{1}{16} \left(-1+i\right)^8-z\right] ^4=\frac{\frac{2}{\sqrt{3}}}{-\frac{1}{\sqrt{3}}+i}$ .

Rješenje.

a)
Sređivanjem lijeve strane zadane jednadžbe slijedi

$\displaystyle \frac{1+7i}{i}$ $\displaystyle =\frac{7-i}{-4}\, z^4, \quad\big /\cdot(-4i)$    
$\displaystyle -4(1+7i)$ $\displaystyle =i(7-i) z^4,$    
$\displaystyle -4(1+7i)$ $\displaystyle =(1+7i) z^4,$    
$\displaystyle -4$ $\displaystyle =z^4.$    

Dakle, trebamo odrediti sve kompleksne brojeve $ z$ za koje vrijedi $ z^4=w$ , gdje je $ w=-4$ . Prema [*] [M1, poglavlje 1.8.1], kompleksni broj $ w$ ima modul $ \vert w\vert=4$ i argument $ \varphi =\pi$ pa je njegov trigonometrijski oblik

$\displaystyle w=4\left(\cos\pi+i\sin\pi\right).$    

Formula [M1, (1.5)] [*] za $ n=4$ daje

$\displaystyle \sqrt[4]{w}=\sqrt[4]{4}\left(\cos\frac{\pi+2k\pi}{4}+\sin\frac{\pi+2k\pi}{4}\right),\quad k=0,1,2,3,$

pa su sva rješenja jednadžbe:

$\displaystyle z_{0}$ $\displaystyle =\sqrt[4]{4}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)= \sqrt{2}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\,i\right)=1+i,$    
$\displaystyle z_{1}$ $\displaystyle =\sqrt[4]{4}\left(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4}\right)= \sqrt{2}\left(-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\,i\right)=-1+i,$    
$\displaystyle z_{2}$ $\displaystyle =\sqrt[4]{4}\left(\cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4}\right)= \sqrt{2}\left(-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\,i\right)=-1-i,$    
$\displaystyle z_{3}$ $\displaystyle =\sqrt[4]{4}\left(\cos\frac{7\pi}{4}+i\sin\frac{7\pi}{4}\right)= \sqrt{2}\left(\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}\,i\right)=1-i.$    

b)
Racionaliziranje desne strane jednadžbe daje

$\displaystyle \frac{\displaystyle\frac{2}{\sqrt{3}}}{-\displaystyle\frac{1}{\sq...
...sqrt{3}}{3}\,i}{-\displaystyle\frac{4}{3}}=
-\frac{1}{2}-\frac{\sqrt{3}}{2}\,i.$

Nadalje, vrijedi

$\displaystyle (-1+i)^8=\left[(-1+i)^2\right]^4=\left(1-2i+i^2\right)^4=\left(-2i\right)^4=(-2)^4i^4=16.$

Uvrštavanjem dobivenih jednakosti u zadanu jednadžbu slijedi

$\displaystyle (1-z)^4=-\frac{1}{2}-\frac{\sqrt{3}}{2}\,i.$

Uz supstituciju $ 1-z=w$ , trebamo riješiti jednadžbu

$\displaystyle w^4=-\frac{1}{2}-\frac{\sqrt{3}}{2}\,i.$

Budući je

$\displaystyle -\frac{1}{2}-\frac{\sqrt{3}}{2}\,i=1\cdot\left(\cos\frac{4\pi}{3}+i \sin\frac{4\pi}{3}\right),$

formula [M1, (1.5)] [*] za $ n=4$ daje

$\displaystyle \sqrt[4]{-\frac{1}{2}-\frac{\sqrt{3}}{2}\,i}=
\sqrt[4]{1}\left(\c...
...+2k\pi}{4}+i \sin\frac{\displaystyle\frac{4\pi}{3}+2k\pi}{4}\right), k=0,1,2,3,$

pa su rješenja:

  $\displaystyle w_0=\cos\frac{\pi}{3}+i \sin\frac{\pi}{3}=\frac{1}{2}+\frac{\sqrt{3}}{2}\,i,$    
  $\displaystyle w_1=\cos\frac{5\pi}{6}+i \sin\frac{5\pi}{6}=-\frac{\sqrt{3}}{2}+\frac{1}{2}\,i,$    
  $\displaystyle w_2=\cos\frac{4\pi}{3}+i \sin\frac{4\pi}{3}=-\frac{1}{2}-\frac{\sqrt{3}}{2}\,i,$    
  $\displaystyle w_3=\cos\frac{11\pi}{6}+i \sin\frac{11\pi}{6}=\frac{\sqrt{3}}{2}-\frac{1}{2}\,i.$    

Kako je $ z=1-w$ , rješenja polazne jednadžbe su:

  $\displaystyle z_0=\frac{1}{2}-\frac{\sqrt{3}}{2}\,i,$    
  $\displaystyle z_1=1+\frac{\sqrt{3}}{2}-\frac{1}{2}\,i,$    
  $\displaystyle z_2=\frac{3}{2}+\frac{\sqrt{3}}{2}\,i,$    
  $\displaystyle z_3=1-\frac{\sqrt{3}}{2}+\frac{1}{2}\,i.$    


Dijeljenje kompleksnih brojeva     OSNOVE MATEMATIKE     Kompleksna ravnina