0 cos \u00b94 sin\u00b9 4 1 2 2 1 cos \u00b9 4 \u00b9 sin \u00b9 4 \u00b9 1 2 2 Third quadrantCos 0 sin 0 C L

# 0 cos ¹4 sin¹ 4 1 2 2 1 cos ¹ 4 ¹ sin ¹ 4 ¹ 1 2

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= 0: ¸ = cos ¹ 4 + ¸ sin ¹ 4 = 1 2 + ¸ 2 . = 1: ¸ = cos ¹ 4 + ¹ + ¸ sin ¹ 4 + ¹ = − 1 2 ¸ 2 . Third quadrant. Cos< 0, sin <0. <>± C L = :;< C L = A/ 4 So the square roots of ¸ are ± 1 2 + ¸ 2
8/12/2015 7 ANOTHER EXAMPLE Example 2: Find all the cube roots of 8. Solution: Since 8 = 2 7 , we know one already: 2 is a root Let’s find the others. Polar form of 8: ’ = 8, ) = 0. 8 = 8. cos · = 8 8 = 1, sin · = 0 8 = 0. So · = 0. So ³ = 8 = 2 7 , · = 0. 8 6 = ³ / 7 cos · 3 + 2 ¹ 3 + ¸ sin · 3 + 2 ¹ 3 = 8 / 7 cos 0 3 + 2 ¹ 3 + ¸ sin 0 3 + 2 ¹ 3 = 2 cos 2 ¹ 3 + ¸ sin 2 ¹ 3 , = 0,1,2. = 0: 8 6 = 2[cos 0 + ¸ sin 0] = 2 1 + 0. ¸ = 2.
8/12/2015 8 8 6 = 2 cos 2 ¹ 3 + ¸ sin 2 ¹ 3 , = 0,1,2. = 1: 8 6 = 2 cos 3 + ¸ sin 3 = 2 cos ¹ − ¹ 3 + ¸ sin ¹ − ¹ 3 = 2 1 2 + ¸ 3 2 = −1 + ¸ 3. = 2: 8 6 = 2 cos 3 + ¸ sin 3 = 2 cos ¹ + ¹ 3 + ¸ sin ¹ + ¹ 3 = 2 1 2 − ¸ 3 2 = −1 − ¸ 3 Second quadrant. Cos< 0, sin >0. <>± C 5 = 5 4 , :;< C 5 = A 4 Third quadrant. Cos< 0, sin <0. FACTORING ³ P − * We calculated : 8 6 = 2, −1 ± ¸ 3. Now if Q = 0, then ³ − ¶ is a factor of Q ³ . (Remainder Theorem) Now ³ 7 − 8 = 0 when ³ = 2, −1 ± 3. So ³ − 2 , ³ + 1 − 3 , and ³ + 1 + 3 Are all the factors of ³ 7 − 8. So ³ 7 − 8 = ³ − 2 ³ + 1 − 3 ³ + 1 + 3 .
8/12/2015 9 STEPS IN FACTORING ³ P − * : Step 1: Write the number * in polar form: * = ´[cos µ + ± sin µ]. Step 2 : Find the n values of *: · * · = * / P \$%& µ ¸ + 2 ° ¸ + ± &±¸ µ ¸ + 2 ° ¸ , = 0,1, … , ¸ − 1. Step 3: If the ¶ values of * · are ³ / , ³ - , … ³ P , Then ³ P − * = ³ − ³ / ³ − ³ - ³ − ³ P . NOW YOU TRY IT: Find the three cube roots of −8. Factorise ³ 7 + 8 into three factors.
8/12/2015 10

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• Summer '20
• Complex number, Imaginary unit

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