Bài tập 23 trang 214 SGK Toán 12 NC

Bài tập 23 trang 214 SGK Toán 12 NC

a) Ta có:

\(\begin{array}{l}
\frac{{4i}}{{1 + i\sqrt 3 }} = \frac{{4i(1 - i\sqrt 3 )}}{4} = \sqrt 3  + i\\
 = 2(\frac{{\sqrt 3 }}{2} + \frac{1}{2}i) = 2(cos\frac{\pi }{6} + {\rm{i}}\sin \frac{\pi }{6})
\end{array}\)

Suy ra: \({(\frac{{4i}}{{1 + i\sqrt 3 }})^6} = {2^6}(\cos \pi  + i\sin \pi ) =  - {2^6}\)

b) Ta có:

\(\begin{array}{l}
{\left( {\sqrt 3  + i} \right)^5} = {2^5}\left( {\cos \frac{{5\pi }}{6} + i\sin \frac{{5\pi }}{6}} \right)\left( 1 \right)\\
1 - i\sqrt 3  = 2\left( {\frac{1}{2} - \frac{{\sqrt 3 }}{2}i} \right)\\
 = 2\left( {\cos \left( { - \frac{\pi }{3}} \right) + i\sin \left( { - \frac{\pi }{3}} \right)} \right)\\
 \Rightarrow {\left( {1 - i\sqrt 3 } \right)^{11}}\\
 = {2^{11}}\left[ {\cos \left( { - \frac{{11\pi }}{3}} \right) + i\sin \left( { - 11\pi 3} \right)} \right]\left( 2 \right)
\end{array}\)

Từ (1) và (2) suy ra:

\(\begin{array}{l}
\frac{{{{(\sqrt 3  + i)}^5}}}{{{{(1 - i\sqrt 3 )}^{11}}}} = \frac{1}{{{2^6}}}[{\rm{cos}}(\frac{{5\pi }}{6} + \frac{{11\pi }}{3})\\
 + {\rm{i}}\sin (\frac{{5\pi }}{6} + \frac{{11\pi }}{3})]\\
 = \frac{1}{{{2^6}}}(cos\frac{{9\pi }}{2} + {\rm{i}}\sin \frac{{9\pi }}{2}) = \frac{i}{{64}}
\end{array}\)$\begin{array}{r}\frac{\mathrm{}}{}\end{array}$

-- Mod Toán 12