by Gọi ${z_1}, {z_2}$ là hai nghiệm của phương trình ${z^2} – 2iz – 4 = 0$, ${z_1}$ có phần thực âm. Tính môđun và acgumen của các số phức sau:
a. $w = z_1^2.{z_2}.$
b. $w = \frac{{{z_1}}}{{{z_2} – 2}}.$
c. $w = \left( {{z_1} – 2} \right)\left( {{z_2} – 2} \right).$
d. $w = \overline {{z_1}.} \left( {2 – \overline {{z_2}} } \right).$
Ta gọi $r$ và $\varphi $ lần lượt là môđun và acgumen của số phức $w.$
Giải phương trình: ${z^2} – 2iz – 4 = 0$ ta được $2$ nghiệm là:
${z_1} = – \sqrt 3 + i = 2\left( { – \frac{{\sqrt 3 }}{2} + \frac{1}{2}i} \right)$ $ = 2\left( {\cos \frac{{5\pi }}{6} + i\sin \frac{{5\pi }}{6}} \right)$ (vì ${z_1}$ có phần thực âm).
${z_2} = \sqrt 3 + i = 2\left( {\frac{{\sqrt 3 }}{2} + \frac{1}{2}i} \right)$ $ = 2\left( {\cos \frac{\pi }{6} + i\sin \frac{\pi }{6}} \right).$
a. Ta có: $z_1^2 = 4\left( {\cos \frac{{5\pi }}{3} + i\sin \frac{{5\pi }}{3}} \right)$, ${z_2} = 2\left( {\cos \frac{\pi }{6} + i\sin \frac{\pi }{6}} \right).$
Suy ra: $w = z_1^2.{z_2}$ $ = 4.2.\left[ {\cos \left( {\frac{{5\pi }}{3} + \frac{\pi }{6}} \right) + i\sin \left( {\frac{{5\pi }}{3} + \frac{\pi }{6}} \right)} \right]$ $ = 8\left( {\cos \frac{{11\pi }}{6} + i\sin \frac{{11\pi }}{6}} \right).$
Vậy $w$ có môđun và một acgumen là: $\left\{ \begin{array}{l}
r = 8\\
\varphi = \frac{{11\pi }}{6}
\end{array} \right.$
b. Ta có
${z_2} – 2 = \sqrt 3 + i – 2$ $ = 2\left( { – 1 + \frac{{\sqrt 3 }}{2} + \frac{1}{2}i} \right)$ $ = 2\left( { – 1 + \cos \frac{\pi }{6} + i\sin \frac{\pi }{6}} \right)$
$ = 2\left( { – 2{{\sin }^2}\frac{\pi }{{12}} + 2i\sin \frac{\pi }{{12}}.\cos \frac{\pi }{{12}}} \right)$ $ = 4\sin \frac{\pi }{{12}}\left( { – \sin \frac{\pi }{{12}} + i\cos \frac{\pi }{{12}}} \right)$
$ = 4\sin \frac{\pi }{{12}}\left[ {\sin \left( { – \frac{\pi }{{12}}} \right) + i\sin \left( { – \frac{\pi }{{12}}} \right)} \right]$ $ = 4\sin \frac{\pi }{{12}}\left( {\cos \frac{{7\pi }}{{12}} + i\sin \frac{{7\pi }}{{12}}} \right).$
Suy ra: $w = \frac{{{z_1}}}{{{z_2} – 2}}$ $ = \frac{{2\left( {\cos \frac{{5\pi }}{6} + i\sin \frac{{5\pi }}{6}} \right)}}{{4\sin \frac{\pi }{{12}}\left( {\cos \frac{{7\pi }}{{12}} + i\sin \frac{{7\pi }}{{12}}} \right)}}$
$ = \frac{1}{{2\sin \frac{\pi }{{12}}}}$$\left[ {\cos \left( {\frac{{5\pi }}{6} – \frac{{7\pi }}{{12}}} \right) + i\sin \left( {\frac{{5\pi }}{6} – \frac{{7\pi }}{{12}}} \right)} \right]$
$ = \frac{1}{{2\sin \frac{\pi }{{12}}}}\left( {\cos \frac{{3\pi }}{{12}} + i\sin \frac{{3\pi }}{{12}}} \right)$ $ = \frac{1}{{2\sin \frac{\pi }{{12}}}}\left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right).$
Vậy $w = \frac{{{z_1}}}{{{z_2} – 2}}$ có môđun và acgumen là $\left\{ \begin{array}{l}
r = \frac{1}{{2\sin \frac{\pi }{{12}}}}\\
\varphi = \frac{\pi }{4}
\end{array} \right.$
c. Ta có ${z_2} – 2$ $ = 4\sin \frac{\pi }{{12}}\left( {\cos \frac{{7\pi }}{{12}} + i\sin \frac{{7\pi }}{{12}}} \right)$ (theo câu b) và:
${z_1} – 2 = – \sqrt 3 + i – 2$ $ = 2\left( { – 1 – \frac{{\sqrt 3 }}{2} + \frac{1}{2}i} \right)$ $ = 2\left( { – 1 – \cos \frac{\pi }{6} + i\sin \frac{\pi }{6}} \right)$
$ = 2\left( { – 2{{\cos }^2}\frac{\pi }{{12}} + 2i\sin \frac{\pi }{{12}}.\cos \frac{\pi }{{12}}} \right)$ $ = 4\cos \frac{\pi }{{12}}\left( { – \cos \frac{\pi }{{12}} + i\sin \frac{\pi }{{12}}} \right)$
$ = 4\cos \frac{\pi }{{12}}\left( {\cos \frac{{11\pi }}{{12}} + i\sin \frac{{11\pi }}{{12}}} \right).$
Suy ra:
$w = \left( {{z_1} – 2} \right)\left( {{z_2} – 2} \right)$ $ = 4\cos \frac{\pi }{{12}}\left( {\cos \frac{{11\pi }}{{12}} + i\sin \frac{{11\pi }}{{12}}} \right)$.$4\sin \frac{\pi }{{12}}\left( {\cos \frac{{7\pi }}{{12}} + i\sin \frac{{7\pi }}{{12}}} \right)$
$ = 16.\sin \frac{\pi }{{12}}.\cos \frac{\pi }{{12}}$$\left[ {\cos \left( {\frac{{11\pi }}{{12}} + \frac{{7\pi }}{{12}}} \right) + i\sin \left( {\frac{{11\pi }}{{12}} + \frac{{7\pi }}{{12}}} \right)} \right]$
$ = 8.\sin \frac{\pi }{6}.\left( {\cos \frac{{18\pi }}{{12}} + i\sin \frac{{18\pi }}{{12}}} \right)$ $ = 8\sin \frac{\pi }{6}\left( {\cos \frac{{3\pi }}{2} + i\sin \frac{{3\pi }}{2}} \right)$
$ = 4\cos \frac{\pi }{{12}}\left( {\cos \frac{{3\pi }}{2} + i\sin \frac{{3\pi }}{2}} \right).$
Vậy $w = \left( {{z_1} – 2} \right)\left( {{z_2} – 2} \right)$ có môđun và một acgumen là: $\left\{ \begin{array}{l}
r = 4\\
\varphi = \frac{{3\pi }}{2}
\end{array} \right.$
Cách khác: Trong trường hợp này, ta có thể áp dụng công thức Vi-et:
${z_1} + {z_2} = 2i, {z_1}{z_2} = – 4.$
Ta có:
$w = \left( {{z_1} – 2} \right)\left( {{z_2} – 2} \right)$ $ = {z_1}.{z_2} – 2\left( {{z_1} + {z_2}} \right) + 4$ $ = – 4 – 2.2i + 4 = – 4i$
$ = 4\left( {0 – i} \right)$ $ = 4\left( {\cos \frac{{3\pi }}{2} + i\sin \frac{{3\pi }}{2}} \right).$
d. $w = \overline {{z_1}} .\left( {2 – \overline {{z_2}} } \right)$ $ \Rightarrow \overline w = \overline {\overline {{z_1}} .\left( {2 – \overline {{z_2}} } \right)} $ $ = {z_1}.\left( {2 – {z_2}} \right) = – {z_1}.\left( {{z_2} – 2} \right)$
Với $ – {z_1} = – 2\left( {\cos \frac{{5\pi }}{6} + i\sin \frac{{5\pi }}{6}} \right)$ $ = 2\left( { – \cos \frac{{5\pi }}{6} – i\sin \frac{{5\pi }}{6}} \right)$
$ = 2\left[ {\cos \left( {\frac{{5\pi }}{6} – \pi } \right) + i\sin \left( {\frac{{5\pi }}{6} – \pi } \right)} \right]$ $ = 2\left[ {\cos \left( { – \frac{\pi }{6}} \right) + i\sin \left( { – \frac{\pi }{6}} \right)} \right]$ và ${z_2} – 2$ $ = 4\sin \frac{\pi }{{12}}\left( {\cos \frac{{7\pi }}{{12}} + i\sin \frac{{7\pi }}{{12}}} \right).$
Suy ra:
$\overline w = – {z_1}.\left( {{z_2} – 2} \right)$ $ = 2\left[ {\cos \left( { – \frac{\pi }{6}} \right) + i\sin \left( { – \frac{\pi }{6}} \right)} \right]$.$4\sin \frac{\pi }{{12}}\left( {\cos \frac{{7\pi }}{{12}} + i\sin \frac{{7\pi }}{{12}}} \right)$
$ = 8\sin \frac{\pi }{{12}}$.$\left[ {\cos \left( { – \frac{\pi }{6} + \frac{{7\pi }}{{12}}} \right) + i\sin \left( { – \frac{\pi }{6} + \frac{{7\pi }}{{12}}} \right)} \right]$ $ = 8\sin \frac{\pi }{{12}}.\left( {\cos \frac{{5\pi }}{{12}} + i\sin \frac{{5\pi }}{{12}}} \right)$
$ \Rightarrow w = 8\sin \frac{\pi }{{12}}$.$\left( {\cos \frac{{5\pi }}{{12}} – i\sin \frac{{5\pi }}{{12}}} \right)$
$ = 8\sin \frac{\pi }{{12}}$.$\left[ {\cos \left( { – \frac{{5\pi }}{{12}}} \right) + i\sin \left( { – \frac{{5\pi }}{{12}}} \right)} \right].$
Vậy $w = \overline {{z_1}} .\left( {2 – \overline {{z_2}} } \right)$ có môđun và acgumen là: $\left\{ \begin{array}{l}
r = 8\sin \frac{\pi }{{12}}\\
\varphi = – \frac{{5\pi }}{{12}}
\end{array} \right.$
