ข้อสอบ A-Level คณิตประยุกต์ 1 ปี 2562 (วิชาสามัญเก่า)

ข้อ 29

ถ้า $z_{1} = \sqrt{2} (\cos \frac{π}{8} + i \sin \frac{π}{8})$ และ $z_{2} = 3 (\cos \frac{3 π}{8} + i \sin \frac{3 π}{8})$ แล้ว $|z_{1} - z_{2}|$ เท่ากับข้อใดต่อไปนี้

1
$\sqrt{5}$
2
$\sqrt{6}$
3
$3$
4
$2 \sqrt{3}$
5
$3 \sqrt{2}$

รีวิว - เสียงตอบรับจากผู้เรียน

เฉลยข้อสอบ

$จากสูตร {|z_{1} - z_{2}|}^{2} = {|z_{1}|}^{2} - z_{1} \bar{z_{2}} - \bar{z_{1}} z_{2} + {|z_{2}|}^{2} \to ⊛ โจทย์ให้ |z_{1}| = \sqrt{2}, |z_{2}| = 3 และ z_{1} = 2 (c o s \frac{π}{8} + i s i n \frac{π}{8}) z_{2} = 3 (c o s \frac{3 π}{8} + i s i n \frac{3 π}{8})$

$จะได้ว่า z_{1} \bar{z_{2}} = [\sqrt{2} (\cos \frac{π}{8} + i \sin \frac{π}{8})] [3 (\cos \frac{3 π}{8} + i \sin \frac{3 π}{8})] จากสูตร z_{A} z_{B} = γ_{A} γ_{B} [\cos (θ_{A} + θ_{B}) + i \sin (θ_{A} + θ_{B})]$

$∴ z_{1} \bar{z_{2}} = [\sqrt{2} (\cos \frac{π}{8} + i \sin \frac{π}{8})] [3 (\cos \frac{3 π}{8} - i \sin (\frac{3 π}{8}))]; \sin (- θ) = - \sin θ = [\sqrt{2} (\cos \frac{π}{8} + i \sin \frac{π}{8})] [3 (\cos (\frac{- 3 π}{8}) + i \sin (\frac{- 3 π}{8}))]; \cos (- θ) = \cos θ$

$= 3 \sqrt{2} [\cos (\frac{π}{8} + (\frac{- 3 π}{8})) + i \sin (\frac{π}{8} + (\frac{- 3 π}{8}))] = 3 \sqrt{2} [\cos (\frac{- π}{4}) + i \sin (\frac{- π}{4})] = 3 \sqrt{2} [\cos (\frac{π}{4}) - i \sin (\frac{π}{4})] = 3 \sqrt{2} [(\frac{1}{\sqrt{2}}) - i (\frac{1}{\sqrt{2}})] = 3 - 3 i$

$∴ {\bar{z}}_{1} z_{2} = [\sqrt{2} (\cos \frac{π}{8} - i \sin \frac{π}{8})] [3 (\cos \frac{3 π}{8} + i \sin \frac{3 π}{8})] = [\sqrt{2} (\cos (\frac{- π}{8}) + i \sin (\frac{- π}{8}))] [3 (\cos \frac{3 π}{8} + i \sin \frac{3 π}{8})]$

$= 3 \sqrt{2} [\cos (\frac{- π}{8} + \frac{3 π}{8}) + i \sin (\frac{- π}{8} + \frac{3 π}{8})] = 3 \sqrt{2} [\cos \frac{π}{4} + i \sin \frac{π}{4}] = 3 \sqrt{2} [\frac{1}{\sqrt{2}} + i (\frac{1}{\sqrt{2}})] = 3 + 3 i$

$จาก ⊛ จะได้ว่า {|z_{1} - z_{2}|}^{2} = {(\sqrt{2})}^{2} - [3 - 3 i] - [3 + 3 i] + {(3)}^{2} = 2 - 3 + 3 i - 3 - 3 i + 9 = 5 ∴ |z_{1} - z_{2}| = \sqrt{5}$

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