Bài 28 trang 205 SGK Giải tích 12 Nâng cao

LG a

\(\eqalign{1 - i\sqrt 3 ;1 + i;(1 - i\sqrt 3 )(1 + i);{{1 - i\sqrt 3 } \over {1 + i}}}\)

Phương pháp giải:

Sử dụng các công thức nhân chia dạng lượng giác của số phức:

\(\begin{array}{l}
{z_1} = {r_1}\left( {\cos {\varphi _1} + i\sin {\varphi _1}} \right)\\
{z_2} = {r_2}\left( {\cos {\varphi _2} + i\sin {\varphi _2}} \right)\\
\Rightarrow {z_1}{z_2} = {r_1}{r_2}\left[ {\cos \left( {{\varphi _1} + {\varphi _2}} \right) + i\sin \left( {{\varphi _1} + {\varphi _2}} \right)} \right]\\
\frac{{{z_1}}}{{{z_2}}} = \frac{{{r_1}}}{{{r_2}}}\left[ {\cos \left( {{\varphi _1} - {\varphi _2}} \right) + i\sin \left( {{\varphi _1} - {\varphi _2}} \right)} \right]
\end{array}\)

Lời giải chi tiết:

\(\eqalign{&1 - i\sqrt 3 = 2\left( {{1 \over 2} - {{\sqrt 3 } \over 2}i} \right) \cr &= 2\left( {\cos \left( { - {\pi \over 3}} \right) + i\sin \left( { - {\pi \over 3}} \right)} \right);\cr 
& 1 + i = \sqrt 2 \left( {{1 \over {\sqrt 2 }} + {1 \over {\sqrt 2 }}i} \right) \cr & = \sqrt 2 \left( {\cos \left( {{\pi \over 4}} \right) + i\sin \left( {{\pi \over 4}} \right)} \right);\, \cr 
& (1 - i\sqrt 3 )(1 + i) \cr & = 2\sqrt 2 \left( {{1 \over 2} - {{\sqrt 3 } \over 2}i} \right)\left( {{1 \over {\sqrt 2 }} + {1 \over {\sqrt 2 }}i} \right) \cr 
& = 2\sqrt 2 \left( {\cos \left( { - {\pi \over 3}} \right) + i\sin \left( { - {\pi \over 3}} \right)} \right)\left( {\cos {\pi \over 4} + i\sin {\pi \over 4}} \right) \cr 
&= 2\sqrt 2 \left[ {\cos \left( {{\pi \over 4} - {\pi \over 3}} \right) + i\sin \left( {{\pi \over 4} - {\pi \over 3}} \right)} \right] \cr 
& = 2\sqrt 2 \left[ {\cos \left( { - {\pi \over {12}}} \right) + i\sin \left( { - {\pi \over {12}}} \right)} \right];\,\, \cr 
& {{1 - i\sqrt 3 } \over {1 + i}} \cr & = \frac{{2\left( {\cos \left( { - \frac{\pi }{3}} \right) + i\sin \left( { - \frac{\pi }{3}} \right)} \right)}}{{\sqrt 2 \left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right)}}\cr &=\frac{2}{{\sqrt 2 }} \left[ {\cos \left( { - {\pi \over 3} - {\pi \over 4}} \right) + i\sin \left( { - {\pi \over 3} - {\pi \over 4}} \right)} \right] \cr 
& = \sqrt 2 \left[ {\cos \left( { - {7 \over {12}}\pi } \right) + i\sin \left( { - {7 \over {12}}\pi } \right)} \right]; \cr} \)

Quảng cáo

LG b

\(\eqalign{2i\left( {\sqrt 3 - i} \right)} \)

Lời giải chi tiết:

\(\eqalign{
& 2i =2\left( {0 + i} \right)= 2\left( {\cos {\pi \over 2} + i\sin {\pi \over 2}} \right) \cr 
&  {\sqrt 3 - i}  = 2\left( {{{\sqrt 3 } \over 2} - {1 \over 2}i} \right) \cr &= 2\left[ {\cos \left( { - {\pi \over 6}} \right) + i\sin \left( { - {\pi \over 6}} \right)} \right]; \cr 
& 2i\left( {\sqrt 3 - i} \right) \cr &= 4\left[ {\cos \left( {{\pi \over 2} - {\pi \over 6}} \right) + i\sin \left( {{\pi \over 2} - {\pi \over 6}} \right)} \right] \cr 
& = 4\left[ {\cos \left( {{\pi \over 3}} \right) + i\sin \left( {{\pi \over 3}} \right)} \right] } \)

Cách khác:

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

LG c

\(\eqalign{{1 \over {2 + 2i}}} \)

Lời giải chi tiết:

\(\eqalign{
& 2 + 2i = 2\sqrt 2 \left( {{1 \over {\sqrt 2 }} + {1 \over {\sqrt 2 }}i} \right) \cr &= 2\sqrt 2 \left( {\cos {\pi \over 4} + i\sin {\pi \over 4}} \right)\, \cr 
& \Rightarrow {1 \over {2 + 2i}} \cr &= {1 \over {2\sqrt 2 }}\left[ {\cos \left( { - {\pi \over 4}} \right) + i\sin \left( { - {\pi \over 4}} \right)} \right] \cr} \)

Cách khác:

\(\begin{array}{l}
\frac{1}{{2 + 2i}} = \frac{{2 - 2i}}{{{2^2} + {2^2}}} = \frac{1}{4} - \frac{1}{4}i\\
= \frac{1}{4}\left( {1 - i} \right)\\
= \frac{1}{4}.\sqrt 2 \left( {\frac{{\sqrt 2 }}{2} - \frac{{\sqrt 2 }}{2}i} \right)\\
= \frac{{\sqrt 2 }}{4}\left( {\cos \left( { - \frac{\pi }{4}} \right) + i\sin \left( { - \frac{\pi }{4}} \right)} \right)
\end{array}\)

LG d

\(\eqalign{z = \sin \varphi + i\cos \varphi \,(\varphi \in\mathbb R)}\)

Lời giải chi tiết:

\(\eqalign{
& z = \,\sin \varphi + i\cos \varphi \cr & =\cos \left( {{\pi \over 2} - \varphi } \right) + i\sin\left( {{\pi \over 2} - \varphi } \right)\cr &(\varphi \in \mathbb R) \cr} \)

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