sqrt(p-8+6i)

[trendafil123]

[tex]\sqrt{ -8+6i}[/tex]

[1089]

kakvo da pomognem?

[Добромир Глухаров]

[tex]-8+6i=r(cos\varphi +i.sin\varphi )[/tex]
[tex]r=\sqrt{(-8)^2+6^2} =\sqrt{64+36}=10[/tex]
[tex]tg\varphi =\frac{6}{-8 } =-\frac{3}{4 }[/tex]
Нека [tex]\varphi_0=\pi -arctg\frac{3}{4 }[/tex]
[tex]\sqrt{-8+6i}=\sqrt{10}\(cos {\frac{\varphi_0+2k\pi }{2 }} +i.sin {\frac{\varphi_0+2k\pi }{2 }}\) ,\ k=0;1[/tex]
[tex]cos\varphi =-0,8[/tex]
[tex]sin\varphi =0,6[/tex]
[tex]cos{\frac{\varphi }{2 }} =\sqrt{\frac{1+cos\varphi }{2 } } =\frac{\sqrt{0,2}}{2}=\frac{\sqrt{10}}{10}[/tex]
[tex]sin{\frac{\varphi }{2 }} =\sqrt{\frac{1-cos\varphi }{2 } } =\sqrt{0,9}=\frac{3}{\sqrt{10}}=\frac{3\sqrt{10}}{10}[/tex]
[tex]cos {\( \frac{\varphi }{2}+\pi \)}=-cos {\( \frac{\varphi }{2 } \)}=-\frac{\sqrt{10}}{10}[/tex]
[tex]sin {\( \frac{\varphi }{2}+\pi \)}=-sin {\( \frac{\varphi }{2 } \)}=-\frac{3\sqrt{10}}{10}[/tex]

Окончателно: [tex]\( \sqrt{-8+6i}\)_1=1+3i,\ \( \sqrt{-8+6i}\)_2=-1-3i[/tex]