سؤال

[SIZE=4]السلام عليكم جد قيمتي X,
Y الحقيقيتين والتي تحقق \[\left ( X+ YI \right )^{5}=\frac{2+2I}{-16I}\]

للسؤال اجابات اخرى

للسؤال اجابات اخرى

\[\begin{array}{l}
{\left( {x + yi} \right)^5} = \frac{{2 + 2i}}{{ - 16i}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ , \Rightarrow {\left( {x + yi} \right)^5} = \frac{{2 + 2i}}{{ - 16i}} \cdot \frac{i}{i} \\
\Rightarrow {\left( {x + yi} \right)^5} = \frac{{ - 1 + i}}{8} = \frac{1}{{\sqrt {32} }}\left( {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right) \\
\Rightarrow \left( {x + yi} \right) = \frac{1}{{\sqrt 2 }}\left( {\cos \frac{{\frac{{3\pi }}{4} + 2k\pi }}{5} + i\sin \frac{{\frac{{3\pi }}{4} + 2k\pi }}{5}} \right)\,\,,\,\,\,\,k = 0,1,2,3,4 \\
\Rightarrow x + y\,i = \frac{1}{{\sqrt 2 }}\left( {\cos \frac{{3\pi + 8k\pi }}{{20}} + i\sin \frac{{3\pi + 8k\pi }}{{20}}} \right)\,\,,\,\,\,\,k = 0,1,2,3,4 \\
If:k = 0 \Rightarrow x + y\,i = \frac{1}{{\sqrt 2 }}\left( {\cos \frac{{3\pi }}{{20}} + i\sin \frac{{3\pi }}{{20}}} \right) \Rightarrow x = \frac{1}{{\sqrt 2 }}\cos \frac{{3\pi }}{{20}}\,\,and\,\,y = \frac{1}{{\sqrt 2 }}\sin \frac{{3\pi }}{{20}} \\
If:k = 1 \Rightarrow x + y\,i = \frac{1}{{\sqrt 2 }}\left( {\cos \frac{{11\pi }}{{20}} + i\sin \frac{{11\pi }}{{20}}} \right) \Rightarrow x = \frac{1}{{\sqrt 2 }}\cos \frac{{11\pi }}{{20}}\,\,and\,\,y = \frac{1}{{\sqrt 2 }}\sin \frac{{11\pi }}{{20}} \\
If:k = 2 \Rightarrow x + y\,i = \frac{1}{{\sqrt 2 }}\left( {\cos \frac{{19\pi }}{{20}} + i\sin \frac{{19\pi }}{{20}}} \right) \Rightarrow x = \frac{1}{{\sqrt 2 }}\cos \frac{{19\pi }}{{20}}\,\,and\,\,y = \frac{1}{{\sqrt 2 }}\sin \frac{{19\pi }}{{20}} \\
If:k = 3 \Rightarrow x + y\,i = \frac{1}{{\sqrt 2 }}\left( {\cos \frac{{27\pi }}{{20}} + i\sin \frac{{27\pi }}{{20}}} \right) \Rightarrow x = \frac{1}{{\sqrt 2 }}\cos \frac{{27\pi }}{{20}}\,\,and\,\,y = \frac{1}{{\sqrt 2 }}\sin \frac{{27\pi }}{{20}} \\
If:k = 4 \Rightarrow x + y\,i = \frac{1}{{\sqrt 2 }}\left( {\cos \frac{{7\pi }}{4} + i\sin \frac{{7\pi }}{4}} \right) = \frac{1}{{\sqrt 2 }}\left( {\frac{1}{{\sqrt 2 }} - \frac{1}{{\sqrt 2 }}i} \right) \Rightarrow \,\,\,\,\,\,x = \frac{1}{2}\,\,\,,\,\,\,\,y = - \frac{1}{2} \\
\end{array}\]