Question

Rozwiąż równanie ze zdjęcia w załączniku

Answer 1

[tex](zi)^4+(1-i)z^2-i^{11}=2i\\z^4+z^2-z^2i+i=2i\\z^4+z^2-z^2i-i=0\\z^2(z^2+1)-i(z^2+1)=0\\(z^2-i)(z^2+1)=0\\z^2-i=0\vee z^2+1=0[/tex]

• [tex]z^2-i=0[/tex]
  
  [tex]z^2-i=0\\(a+bi)^2-i=0\\a^2+2abi-b^2-i=0\\a^2-b^2+(2ab-1)i=0\\a^2-b^2=-(2ab-1)i\\\Downarrow\\a^2-b^2=0 \wedge -(2ab-1)=0\\\\a^2-b^2=0\\(a-b)(a+b)=0\\a=b \vee a=-b\\\\-(2b^2-1)=0\\2b^2-1=0\\2b^2=1\\b^2=\dfrac{1}{2}\\b=\dfrac{\sqrt2}{2} \vee b=-\dfrac{\sqrt{2}}{2}\\\\-(-2b^2-1)=0\\2b^2+1=0\\2b^2=-1\\b^2=-\dfrac{1}{2}\\b\in\emptyset\\\\a=\dfrac{\sqrt2}{2} \vee a=-\dfrac{\sqrt2}{2}\\\\z=\dfrac{\sqrt2}{2}+\dfrac{\sqrt2}{2}i \vee z=\dfrac{\sqrt2}{2}-\dfrac{\sqrt2}{2}i[/tex]

• [tex]z^2+1=0[/tex]
  
  [tex]z^2+1=0\\z^2=-1\\z=\sqrt{-1} \vee z=-\sqrt{-1}\\z=i \vee z=-i[/tex]

[tex]z\in\left\{\dfrac{\sqrt2}{2}-\dfrac{\sqrt2}{2}i,\dfrac{\sqrt2}{2}+\dfrac{\sqrt2}{2}i,-i,i\right\}[/tex]