Viết số phức dưới dạng lượng giác – https://mindovermetal.org

Phương pháp
Để viết số phức $z = a + bi,(a,b \in R)$ dưới dạng lượng giác $z = r(c{\rm{os}}\varphi + i\sin \varphi )$, trước hết ta biến đổi: $z = \sqrt {{a^2} + {b^2}} (\frac{a}{{\sqrt {{a^2} + {b^2}} }} + \frac{b}{{\sqrt {{a^2} + {b^2}} }}i).$
Như vậy: $r = \sqrt {{a^2} + {b^2}}.$ Đặt $c{\rm{os}}\varphi = \frac{a}{{\sqrt {{a^2} + {b^2}} }}$ và $\sin \varphi = \frac{b}{{\sqrt {{a^2} + {b^2}} }}.$
Từ đó suy ra $\varphi $ là $1$ $acgumen$ của $z.$

Các công thức biến đổi lượng giác cần lưu ý

$1 + c{\rm{os}}\varphi + i\sin \varphi $ $ = 2{\cos ^2}\frac{\varphi }{2} + 2i\sin \frac{\varphi }{2}c{\rm{os}}\frac{\varphi }{2}$ $ = 2\cos \frac{\varphi }{2}\left[ {c{\rm{os}}\frac{\varphi }{2} + i \sin \frac{\varphi }{2}} \right].$
$1 + i\tan \varphi = 1 + i\frac{{\sin \varphi }}{{c{\rm{os}}\varphi }}$ $ = \frac{1}{{c{\rm{os}}\varphi }}(c{\rm{os}}\varphi + i \sin \varphi ).$

Ví dụ 1. Viết các số phức sau dưới dạng lượng giác:
a. $5$.
b. $-3$.
c. $7i$.
d. $-2i$.

a. $5 = 5\left( {1 + 0i} \right) = 5\left( {\cos 0 + i\sin 0} \right).$
b. $ – 3 = 3\left( { – 1 + 0i} \right) = 3\left( {{\rm{cos}}\pi {\rm{ + sin}}\pi {\rm{i}}} \right).$
c. $7i = 7\left( {0 + i} \right) = 7\left( {\cos \frac{\pi }{2} + i\sin \frac{\pi }{2}} \right).$
d. $ – 2i = 2\left( {0 – i} \right)$ $ = 2\left( {\cos \left( { – \frac{\pi }{2}} \right) + i\sin \left( { – \frac{\pi }{2}} \right)} \right).$

Ví dụ 2. Viết các số phức sau dưới dạng lượng giác:
a. $1 – i\sqrt 3.$
b. $\sqrt 3 – i\sqrt 3 .$
c. $\frac{1}{3} + \frac{{\sqrt 3 }}{3}i.$
d. $\frac{{7\sqrt 3 }}{3} – 7i.$

a. $1 – i\sqrt 3 = 2\left( {\frac{1}{2} – i\frac{{\sqrt 3 }}{2}} \right)$ $ = 2\left[ {\cos \left( { – \frac{\pi }{3}} \right) + i\sin \left( { – \frac{\pi }{3}} \right)} \right].$
b. $\sqrt 3 – i\sqrt 3 = \sqrt 3 \left( {1 – i} \right)$ $ = \sqrt 6 \left( {\frac{1}{{\sqrt 2 }} – \frac{i}{{\sqrt 2 }}} \right)$ $ = \sqrt 6 \left[ {\cos \left( { – \frac{\pi }{4}} \right) + i\sin \left( { – \frac{\pi }{4}} \right)} \right].$
c. $\frac{1}{3} + \frac{{\sqrt 3 }}{3}i = \frac{2}{3}\left( {\frac{1}{2} + i\frac{{\sqrt 3 }}{2}} \right)$ $ = \frac{2}{3}\left( {\cos \frac{\pi }{3} + i\sin \frac{\pi }{3}} \right).$
d. $\frac{{7\sqrt 3 }}{3} – 7i = \frac{{7\sqrt 3 }}{3}\left( {1 – i\sqrt 3 } \right)$ $ = \frac{{14\sqrt 3 }}{3}\left( {\frac{1}{2} – i\frac{{\sqrt 3 }}{2}} \right)$ $ = \frac{{14\sqrt 3 }}{3}\left[ {\cos \left( { – \frac{\pi }{3}} \right) + i\sin \left( { – \frac{\pi }{3}} \right)} \right].$

Ví dụ 3. Viết các số phức sau dưới dạng lượng giác:
a. $\left( {1 + 3i} \right)\left( {1 + 2i} \right).$
b. $\left( {1 + i} \right)\left[ {1 + \left( {\sqrt 3 – 2} \right)i} \right].$
c. $\left( {\sqrt 2 – 2i} \right).\left[ {\sqrt 2 + \left( {3\sqrt 2 – 4} \right)i} \right].$

a. $\left( {1 + 3i} \right)\left( {1 + 2i} \right)$ $ = 1 + 6{i^2} + 3i + 2i$ $ = – 5 + 5i = 5\left( { – 1 + i} \right)$
$ = 5\sqrt 2 \left( { – \frac{1}{{\sqrt 2 }} + i\frac{1}{{\sqrt 2 }}} \right)$ $ = 5\sqrt 2 \left( {\cos \frac{{3\pi }}{4} + i\sin \frac{{3\pi }}{4}} \right).$
b. $\left( {1 + i} \right)\left[ {1 + \left( {\sqrt 3 – 2} \right)i} \right]$ $ = 1 – \left( {\sqrt 3 – 2} \right) + \left( {\sqrt 3 – 2 + 1} \right)i$
$ = 3 – \sqrt 3 + \left( {\sqrt 3 – 1} \right)i$ $ = \sqrt 3 \left( {\sqrt 3 – 1} \right) + \left( {\sqrt 3 – 1} \right)i$
$ = \left( {\sqrt 3 – 1} \right)\left( {\sqrt 3 + i} \right)$ $ = 2\left( {\sqrt 3 – 1} \right)\left( {\frac{{\sqrt 3 }}{2} + \frac{1}{2}i} \right)$ $ = 2\left( {\sqrt 3 – 2} \right)\left( {\cos \frac{\pi }{6} + i\sin \frac{\pi }{6}} \right).$
c. $\left( {\sqrt 2 – 2i} \right).\left[ {\sqrt 2 + \left( {3\sqrt 2 – 4} \right)i} \right]$ $ = \left( {2 + 6\sqrt 2 – 8} \right) + \left( {6 – 4\sqrt 2 – 2\sqrt 2 } \right)i$
$ = \left( {6\sqrt 2 – 6} \right) + \left( {6 – 6\sqrt 2 } \right)i$ $ = \left( {6\sqrt 2 – 6} \right)\left( {1 – i} \right)$
$ = \sqrt 2 \left( {6\sqrt 2 – 6} \right)\left( {\frac{1}{{\sqrt 2 }} – \frac{1}{{\sqrt 2 }}i} \right)$ $ = \left( {12 – 6\sqrt 2 } \right)\left[ {\cos \left( { – \frac{\pi }{4}} \right) + i\sin \left( { – \frac{\pi }{4}} \right)} \right].$

Ví dụ 4. Viết các số phức sau dưới dạng lượng giác:
a. $\frac{1}{{2 + 2i}}.$
b. $\frac{{3 – i}}{{1 – 2i}}.$
c. $\frac{{1 – i\sqrt 3 }}{{1 + i}}.$

a. Ta có:
$\frac{1}{{2 + 2i}} = \frac{1}{{2\left( {1 + i} \right)}}$ $ = \frac{{\sqrt 2 }}{{4\left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right)}}$ $ = \frac{{\sqrt 2 }}{4}\left[ {\cos \left( { – \frac{\pi }{4}} \right) + i\sin \left( { – \frac{\pi }{4}} \right)} \right].$
b. $\frac{{3 – i}}{{1 – 2i}} = \frac{{\left( {3 – i} \right)\left( {1 + 2i} \right)}}{{\left( {1 – 2i} \right)\left( {1 + 2i} \right)}}$ $ = \frac{{3 + 2 + 6i – i}}{{1 – {{\left( {2i} \right)}^2}}} = \frac{{5 + 5i}}{{1 + 4}}$ $ = 1 + i$
$ = \sqrt 2 \left( {\frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 2 }}i} \right)$ $ = \sqrt 2 \left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right).$
c. $\frac{{1 – i\sqrt 3 }}{{1 + i}}$ $ = \frac{2}{{\sqrt 2 }}\left[ {\cos \left( { – \frac{\pi }{3} – \frac{\pi }{4}} \right) + i\sin \left( { – \frac{\pi }{3} – \frac{\pi }{4}} \right)} \right]$ $ = \sqrt 2 \left[ {\cos \left( {\frac{{ – 7\pi }}{{12}}} \right) + i\sin \left( {\frac{{ – 7\pi }}{{12}}} \right)} \right].$

Ví dụ 5. Viết các số phức sau dưới dạng lượng giác:
a. $1 + \frac{i}{{\sqrt 3 }}.$
b. $1 + \sqrt 3 + \left( {1 – \sqrt 3 } \right)i.$

a. Ta có:
$1 + \frac{i}{{\sqrt 3 }} = 1 + i\tan \frac{\pi }{6}$ $ = 1 + i\frac{{\sin \frac{\pi }{6}}}{{\cos \frac{\pi }{6}}}$ $ = \frac{1}{{\cos \frac{\pi }{6}}}\left( {\cos \frac{\pi }{6} + i\sin \frac{\pi }{6}} \right)$ $ = \frac{2}{{\sqrt 3 }}\left( {\cos \frac{\pi }{6} + i\sin \frac{\pi }{6}} \right).$
b. $1 + \sqrt 3 + \left( {1 – \sqrt 3 } \right)i$ $ = 1 + \tan \frac{\pi }{3} + \left( {1 – \tan \frac{\pi }{3}} \right)i$ $ = 1 + \frac{{\sin \frac{\pi }{3}}}{{\cos \frac{\pi }{3}}} + \left( {1 – \frac{{\sin \frac{\pi }{3}}}{{\cos \frac{\pi }{3}}}} \right)i$
$ = \frac{1}{{\cos \frac{\pi }{3}}}\left( {\cos \frac{\pi }{3} + \sin \frac{\pi }{3}} \right)$ $ + \frac{1}{{\cos \frac{\pi }{3}}}\left( {\cos \frac{\pi }{3} + i\sin \frac{\pi }{3}} \right)i$ $ = \frac{1}{{\cos \frac{\pi }{3}}}\left( {\cos \frac{\pi }{3} + \sin \frac{\pi }{3}} \right)$ $ – \frac{1}{{\cos \frac{\pi }{3}}}\left( {\sin \frac{\pi }{3} – \cos \frac{\pi }{3}} \right)i$
$ = \frac{1}{{\cos \frac{\pi }{3}}}\sqrt 2 \cos \left( {\frac{\pi }{3} – \frac{\pi }{4}} \right)$ $ – \frac{1}{{\cos \frac{\pi }{3}}}\sqrt 2 \sin \left( {\frac{\pi }{3} – \frac{\pi }{4}} \right).i$
$ = 2\sqrt 2 \left( {\cos \frac{\pi }{{12}} – i\sin \frac{\pi }{{12}}} \right)$ $ = 2\sqrt 2 \left[ {\cos \left( { – \frac{\pi }{{12}}} \right) + i\sin \left( { – \frac{\pi }{{12}}} \right)} \right].$
Cách khác:
$1 + \sqrt 3 + \left( {1 – \sqrt 3 } \right)i$ $ = \left( {1 + \sqrt 3 } \right)\left( {1 + \frac{{1 – \sqrt 3 }}{{1 + \sqrt 3 }}i} \right)$ $ = \left( {1 + \sqrt 3 } \right)\left( {1 + \frac{{\tan \frac{\pi }{4} – \tan \frac{\pi }{3}}}{{1 + \tan \frac{\pi }{4}.\tan \frac{\pi }{3}}}i} \right)$
$ = \left( {1 + \sqrt 3 } \right)\left[ {1 + i\tan \left( {\frac{\pi }{4} – \frac{\pi }{3}} \right)} \right]$ $ = \left( {1 + \sqrt 3 } \right)\left[ {1 + i\tan \left( { – \frac{\pi }{{12}}} \right)} \right]$
$ = \left( {1 + \sqrt 3 } \right)\left[ {1 + i\frac{{\sin \left( { – \frac{\pi }{{12}}} \right)}}{{\cos \left( { – \frac{\pi }{{12}}} \right)}}} \right]$ $ = \frac{{1 + \sqrt 3 }}{{\cos \frac{\pi }{{12}}}}\left[ {\cos \left( { – \frac{\pi }{{12}}} \right) + i\sin \left( { – \frac{\pi }{{12}}} \right)} \right].$
Mà $\cos \frac{\pi }{{12}} = \cos \left( {\frac{\pi }{3} – \frac{\pi }{4}} \right)$ $ = \cos \frac{\pi }{3}.\cos \frac{\pi }{4} + \sin \frac{\pi }{3}.\sin \frac{\pi }{4}$ $ = \frac{1}{{2\sqrt 2 }} + \frac{{\sqrt 3 }}{{2\sqrt 2 }} = \frac{{1 + \sqrt 3 }}{{2\sqrt 2 }}.$
Do đó: $1 + \sqrt 3 + \left( {1 – \sqrt 3 } \right).i$ $ = \frac{{1 + \sqrt 3 }}{{\cos \frac{\pi }{{12}}}}\left[ {\cos \left( { – \frac{\pi }{{12}}} \right) + i\sin \left( { – \frac{\pi }{{12}}} \right)} \right]$ $ = 2\sqrt 2 \left[ {\cos \left( { – \frac{\pi }{{12}}} \right) + i\sin \left( { – \frac{\pi }{{12}}} \right)} \right].$

Source: https://mindovermetal.org
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