ข้อสอบ PAT 1 - มีนาคม 2564

$$\begin{gathered} \mathrm{จากสูตร} z = rcis\theta \\ \mathrm{จะได้} z^n = r^{n}cisn\theta \\ z^{\frac{1}{n}} = r^{\frac{1}{n}}cis\left(\frac{\theta +2k\mathrm{\pi }}{n}\right) \\ \mathrm{โดย} k=0, 1, 2, \dots , n-1 \end{gathered}$$

$$\begin{gathered} \mathrm{จากโจทย์} z_1 = \cos \left(\frac{\mathrm{\pi }}{5}\right)+i\sin \left(\frac{\mathrm{\pi }}{5}\right) \\ \mathrm{เป็นรากที่สิบของจำนวนเชิงซ้อนจำนวนหนึ่ง} \\ \mathrm{จะได้} z_1 = z^{\frac{1}{10}} \\ z^{10} = z \to z=z^{10} \\ z = {\left(\cos \frac{\mathrm{\pi }}{5}+i\sin \frac{\mathrm{\pi }}{5}\right)}^{10} \\ z = {\left(cis\frac{\mathrm{\pi }}{5}\right)}^{10} = 1^{10}cis\frac{10\mathrm{\pi }}{5} \\ z = cis2\mathrm{\pi } ; \mathrm{r}=1 , \mathrm{\theta }=2\mathrm{\pi } \end{gathered}$$

$$\begin{gathered} \mathrm{โดย} z=cis2\mathrm{\pi } \mathrm{มีรากทั้งหมด} 10 \mathrm{ราก} ({\mathrm{z}}_1, {\mathrm{z}}_2, \dots , {\mathrm{z}}_{10}) \\ \mathrm{จากสูตร} {\mathrm{z}}^{\frac{1}{\mathrm{n}}} = {\mathrm{r}}^{\frac{1}{\mathrm{n}}} \mathrm{cis}\left(\frac{\mathrm{\theta }+2\mathrm{k\pi }}{\mathrm{n}}\right) \end{gathered}$$

$$\begin{gathered} \mathrm{k}=0 {\mathrm{z}}_1 = {\mathrm{z}}^{\frac{1}{10}} = 1^{\frac{1}{10}}\mathrm{cis}\left(\frac{2\mathrm{\pi }+0\mathrm{\pi }}{10}\right) = \mathrm{cis}\frac{\mathrm{\pi }}{5} = \cos \frac{\mathrm{\pi }}{5}+\mathrm{isin}\frac{\mathrm{\pi }}{5} \\ \mathrm{k}=1 {\mathrm{z}}_2 = {\mathrm{z}}^{\frac{1}{10}} = 1^{\frac{1}{10}}\mathrm{cis}\left(\frac{2\mathrm{\pi }+2\left(1\right)\mathrm{\pi }}{10}\right) = \mathrm{cis}\frac{2\mathrm{\pi }}{5} = \cos \frac{2\mathrm{\pi }}{5}+\mathrm{isin}\frac{2\mathrm{\pi }}{5} \\ \mathrm{k}=2 {\mathrm{z}}_3 = {\mathrm{z}}^{\frac{1}{10}} = 1^{\frac{1}{10}}\mathrm{cis}\left(\frac{2\mathrm{\pi }+2\left(2\right)\mathrm{\pi }}{10}\right) = \mathrm{cis}\frac{3\mathrm{\pi }}{5} = \cos \frac{3\mathrm{\pi }}{5}+\mathrm{isin}\frac{3\mathrm{\pi }}{5} \end{gathered}$$
$$\begin{gathered} \mathrm{k}=3 {\mathrm{z}}_4 = {\mathrm{z}}^{\frac{1}{10}} = 1^{\frac{1}{10}}\mathrm{cis}\left(\frac{2\mathrm{\pi }+2\left(3\right)\mathrm{\pi }}{10}\right) = \mathrm{cis}\frac{4\mathrm{\pi }}{5} = \cos \frac{4\mathrm{\pi }}{5}+\mathrm{isin}\frac{4\mathrm{\pi }}{5} \\ \mathrm{k}=4 {\mathrm{z}}_5 = {\mathrm{z}}^{\frac{1}{10}} = 1^{\frac{1}{10}}\mathrm{cis}\left(\frac{2\mathrm{\pi }+2\left(4\right)\mathrm{\pi }}{10}\right) = \mathrm{cis\pi } = \mathrm{cos\pi }+\mathrm{isin\pi } \\ \mathrm{k}=5 {\mathrm{z}}_6 = {\mathrm{z}}^{\frac{1}{10}} = 1^{\frac{1}{10}}\mathrm{cis}\left(\frac{2\mathrm{\pi }+2\left(5\right)\mathrm{\pi }}{10}\right) = \mathrm{cis}\frac{6\mathrm{\pi }}{5} = \cos \frac{6\mathrm{\pi }}{5}+\mathrm{isin}\frac{6\mathrm{\pi }}{5} \end{gathered}$$
$$\begin{gathered} \mathrm{k}=6 {\mathrm{z}}_7 = {\mathrm{z}}^{\frac{1}{10}} = 1^{\frac{1}{10}}\mathrm{cis}\left(\frac{2\mathrm{\pi }+2\left(6\right)\mathrm{\pi }}{10}\right) = \mathrm{cis}\frac{7\mathrm{\pi }}{5} = \cos \frac{7\mathrm{\pi }}{5}+\mathrm{isin}\frac{7\mathrm{\pi }}{5} \\ \mathrm{k}=7 {\mathrm{z}}_8 = {\mathrm{z}}^{\frac{1}{10}} = 1^{\frac{1}{10}}\mathrm{cis}\left(\frac{2\mathrm{\pi }+2\left(7\right)\mathrm{\pi }}{10}\right) = \mathrm{cis}\frac{8\mathrm{\pi }}{5} = \cos \frac{8\mathrm{\pi }}{5}+\mathrm{isin}\frac{8\mathrm{\pi }}{5} \\ \mathrm{k}=8 {\mathrm{z}}_9 = {\mathrm{z}}^{\frac{1}{10}} = 1^{\frac{1}{10}}\mathrm{cis}\left(\frac{2\mathrm{\pi }+2\left(8\right)\mathrm{\pi }}{10}\right) = \mathrm{cis}\frac{9\mathrm{\pi }}{5} = \cos \frac{9\mathrm{\pi }}{5}+\mathrm{isin}\frac{9\mathrm{\pi }}{5} \end{gathered}$$
$$\begin{gathered} \mathrm{k}=9 {\mathrm{z}}_{10} = {\mathrm{z}}^{\frac{1}{10}} = 1^{\frac{1}{10}}\mathrm{cis}\left(\frac{2\mathrm{\pi }+2\left(9\right)\mathrm{\pi }}{10}\right) = \mathrm{cis}2\mathrm{\pi } = \cos 2\mathrm{\pi }+\mathrm{isin}2\mathrm{\pi } \\ = \cos 0+\mathrm{isin}0 \end{gathered}$$

$$\begin{gathered} \mathrm{ดังนั้นข้อที่ไม่ใช้รากที่สิบของจำนวนเชิงซ้อนของโจทย์คือ} \\ \mathrm{ข้อ} 2. \cos \left(\frac{\mathrm{\pi }}{2}\right)+i\sin \left(\frac{\mathrm{\pi }}{2}\right) \end{gathered}$$