ข้อสอบ PAT 1 - มีนาคม 2564

$จากสูตร z = r c i s θ จะได้ z^{n} = r^{n} c i s n θ z^{\frac{1}{n}} = r^{\frac{1}{n}} c i s (\frac{θ + 2 k π}{n}) โดย k = 0, 1, 2, \dots, n - 1$

$จากโจทย์ z_{1} = \cos (\frac{π}{5}) + i \sin (\frac{π}{5}) เป็นรากที่สิบของจำนวนเชิงซ้อนจำนวนหนึ่ง จะได้ z_{1} = z^{\frac{1}{10}} z^{10} = z \to z = z^{10} z = {(\cos \frac{π}{5} + i \sin \frac{π}{5})}^{10} z = {(c i s \frac{π}{5})}^{10} = 1^{10} c i s \frac{10 π}{5} z = c i s 2 π; r = 1, θ = 2 π$

$โดย z = c i s 2 π มีรากทั้งหมด 10 ราก (z_{1}, z_{2}, \dots, z_{10}) จากสูตร z^{\frac{1}{n}} = r^{\frac{1}{n}} cis (\frac{θ + 2 kπ}{n}) k = 0 z_{1} = z^{\frac{1}{10}} = 1^{\frac{1}{10}} cis (\frac{2 π + 0 π}{10}) = cis \frac{π}{5} = \cos \frac{π}{5} + isin \frac{π}{5} k = 1 z_{2} = z^{\frac{1}{10}} = 1^{\frac{1}{10}} cis (\frac{2 π + 2 (1) π}{10}) = cis \frac{2 π}{5} = \cos \frac{2 π}{5} + isin \frac{2 π}{5} k = 2 z_{3} = z^{\frac{1}{10}} = 1^{\frac{1}{10}} cis (\frac{2 π + 2 (2) π}{10}) = cis \frac{3 π}{5} = \cos \frac{3 π}{5} + isin \frac{3 π}{5}$

$k = 3 z_{4} = z^{\frac{1}{10}} = 1^{\frac{1}{10}} cis (\frac{2 π + 2 (3) π}{10}) = cis \frac{4 π}{5} = \cos \frac{4 π}{5} + isin \frac{4 π}{5} k = 4 z_{5} = z^{\frac{1}{10}} = 1^{\frac{1}{10}} cis (\frac{2 π + 2 (4) π}{10}) = cisπ = cosπ + isinπ k = 5 z_{6} = z^{\frac{1}{10}} = 1^{\frac{1}{10}} cis (\frac{2 π + 2 (5) π}{10}) = cis \frac{6 π}{5} = \cos \frac{6 π}{5} + isin \frac{6 π}{5}$

$k = 6 z_{7} = z^{\frac{1}{10}} = 1^{\frac{1}{10}} cis (\frac{2 π + 2 (6) π}{10}) = cis \frac{7 π}{5} = \cos \frac{7 π}{5} + isin \frac{7 π}{5} k = 7 z_{8} = z^{\frac{1}{10}} = 1^{\frac{1}{10}} cis (\frac{2 π + 2 (7) π}{10}) = cis \frac{8 π}{5} = \cos \frac{8 π}{5} + isin \frac{8 π}{5} k = 8 z_{9} = z^{\frac{1}{10}} = 1^{\frac{1}{10}} cis (\frac{2 π + 2 (8) π}{10}) = cis \frac{9 π}{5} = \cos \frac{9 π}{5} + isin \frac{9 π}{5}$