Question

sa se rezolve ecuatia :$(x+i)^n+(x-i)^n=0$

Answer 1

[tex](x+i)^n+(x-i)^n=0\\ Pt. x=i\ \text{se obtine:}\\ 2^n\cdot i^n=0,\text{ care nu are solutii.}\\ Pt. x\neq i\ \text{avem:}\\ (x+i)^n=-(x-i)^n\\ \left(\dfrac{x+i}{x-i}\right)^n=-1\\ \text{Daca scriem sub forma trigonometrica vom avea:}\\ \left(\dfrac{x+i}{x-i}\right)^n= \cos \pi +i\cdot \sin \pi\\ \dfrac{x+i}{x-i}= \cos \dfrac{\pi +2k\pi}{n}+i\sin \dfrac{\pi +2k\pi}{n},k=\overline{0,n-1}\\ Notam: y_k= \cos \dfrac{\pi +2k\pi}{n}+i\sin \dfrac{\pi +2k\pi}{n}\\\ \dfrac{x+i}{x-i}=y_k\\ x+i=(x-i)\cdot y_k\\ [/tex]
[tex]x+i=x\cdot y_k-i\cdot y_k\\ x(1-y_k)=i(1+y_k)\\ x=\dfrac{i+i\cdot y_k}{1-y_k}\\ x=\dfrac{i(1+\cos \dfrac{\pi +2k\pi}{n}+i\sin \dfrac{\pi +2k\pi}{n})}{1- \cos \dfrac{\pi +2k\pi}{n}-i\sin \dfrac{\pi +2k\pi}{n}}\\ \text{Mai departe daca stii formulele:}\\ \bullet1+\cos t=2\cos ^2\dfrac{t}{2},\ 1-\cos t=2\ sin^2 \dfrac{t}{2}\\ \bullet 1+\cos t+i\cdot \sin t=2\cos^2 \dfrac{t}{2}+2i\sin \dfrac{t}{2}\cdot \cos \dfrac{t}{2}=\\ =2\cdot \cos \dfrac{t}{2}\left(cos \dfrac{t}{2}+i\sin \dfrac{t}{2}\right)\\ [/tex]
[tex]\bullet 1-\cos t-i\sin t=2\sin^2 \dfrac{t}{2}-2i\sin \dfrac{t}{2}\cos \dfrac{t}{2}=\\ =-2i\left(\cos \dfrac{t}{2}+i\sin \dfrac{t}{2}\right)\\ Revenind:\\ x_k=\dfrac{i\cdot 2\cos \dfrac{\pi+2k\pi}{2n}\left(\cos \dfrac{\pi+2k\pi}{2n}+i\cdot \sin \dfrac{\pi+2k\pi}{2n}\right)}{-2i\cdot \sin \dfrac{\pi+2k\pi}{2n}\left(\cos \dfrac{\pi+2k\pi}{2n}+i\cdot \sin \dfrac{\pi+2k\pi}{2n}\right)} }\\ x_k=-\dfrac{\cos \dfrac{\pi+2k\pi}{2n}}{\sin \dfrac{\pi+2k\pi}{2n}}\\ \boxed{x_k=-ctg \dfrac{\pi+2k\pi}{2n}}.\\ [/tex]
$\text{Asa arata cele n-1 solutii ale ecuatiei.}$