Question 1 How do you simplify 2 ( cos ⁡ 24 0 ∘ + 𝑖 sin ⁡ 24 0 ∘ ) ⋅ 3 ( cos ⁡ 6 0 ∘ + 𝑖 sin ⁡ 6 0 ∘ ) 2(cos240 ∘ +isin240 ∘ )⋅3(cos60 ∘ +isin60 ∘ ) and express the result in rectangular form? Answer 1 Answer: 3 − 3 𝑖 3 3−3i 3 Explanation: First, find the values of the trigonometric functions: cos ⁡ 24 0 ∘ = − 1 2 , sin ⁡ 24 0 ∘ = − 3 2 cos240 ∘ =− 2 1 ,sin240 ∘ =− 2 3 cos ⁡ 6 0 ∘ = 1 2 , sin ⁡ 6 0 ∘ = 3 2 cos60 ∘ = 2 1 ,sin60 ∘ = 2 3 Use 𝑖 2 = − 1 i 2 =−1: 2 ( cos ⁡ 24 0 ∘ + 𝑖 sin ⁡ 24 0 ∘ ) ⋅ 3 ( cos ⁡ 6 0 ∘ + 𝑖 sin ⁡ 6 0 ∘ ) = 6 ( − 1 2 − 𝑖 3 2 ) ( 1 2 + 𝑖 3 2 ) 2(cos240 ∘ +isin240 ∘ )⋅3(cos60 ∘ +isin60 ∘ )=6(− 2 1 −i 2 3 )( 2 1 +i 2 3 ) Expand and simplify: = 6 ( − 1 4 − 𝑖 3 4 − 𝑖 3 4 − 3 4 ) = 3 − 3 𝑖 3 =6(− 4 1 −i 4 3 −i 4 3 − 4 3 )=3−3i 3 Question 2 How do you find the power ( − 2 + 2 𝑖 ) 3 (−2+2i) 3 and express the result in rectangular form? Answer 1 Convert − 2 + 2 𝑖 −2+2i to polar form: Magnitude 𝑟 = 2 2 r=2 2 Argument 𝜃 = 3 𝜋 4 θ= 4 3π Calculate ( − 2 + 2 𝑖 ) 3 (−2+2i) 3 : ( − 2 + 2 𝑖 ) 3 = ( 2 2 𝑒 𝑖 3 𝜋 4 ) 3 = 16 ( 1 + 𝑖 ) (−2+2i) 3 =(2 2 e i 4 3π ) 3 =16(1+i) So the answer is 16 ( 1 + 𝑖 ) 16(1+i). Question 3 How do you find the cube root of 81 ( cos ⁡ 𝜋 12 + 𝑖 sin ⁡ 𝜋 12 ) 81(cos 12 π +isin 12 π )? Answer 1 Write 81 ( cos ⁡ 𝜋 12 + 𝑖 sin ⁡ 𝜋 12 ) 81(cos 12 π +isin 12 π ) as 81 cis 𝜋 12 81cis 12 π . Taking the cube root: ( 81 cis 𝜋 12 ) 1 3 = 3 cis ( 𝜋 12 × 1 3 ) (81cis 12 π ) 3 1 =3cis( 12 π × 3 1 ) This gives the cube roots at angles of 1 5 ∘ 15 ∘ , 12 5 ∘ 125 ∘ , and 24 5 ∘ 245 ∘ : 3 ( cos ⁡ 1 5 ∘ + 𝑖 sin ⁡ 1 5 ∘ ) , 3 ( cos ⁡ 12 5 ∘ + 𝑖 sin ⁡ 12 5 ∘ ) , 3 ( cos ⁡ 24 5 ∘ + 𝑖 sin ⁡ 24 5 ∘ ) 3(cos15 ∘ +isin15 ∘ ),3(cos125 ∘ +isin125 ∘ ),3(cos245 ∘ +isin245 ∘ ) Question 4 How can I calculate the square root of 5 ( cos ⁡ 12 0 ∘ + 𝑖 sin ⁡ 12 0 ∘ ) 5(cos120 ∘ +isin120 ∘ )? Answer 1 Convert to polar form and take the square root: 5 ( cos ⁡ 6 0 ∘ + 𝑖 sin ⁡ 6 0 ∘ ) = 5 2 ( 1 + 𝑖 3 ) 5 (cos60 ∘ +isin60 ∘ )= 2 5 (1+i 3 ) Question 5 What does ( 1 + 𝑖 ) × ( 6 − 2 𝑖 ) 4 𝑖 4i (1+i)×(6−2i) equal? Answer 1 Expand the numerator: ( 1 + 𝑖 ) ( 6 − 2 𝑖 ) = 6 − 2 𝑖 + 6 𝑖 − 2 𝑖 2 = 8 + 4 𝑖 (1+i)(6−2i)=6−2i+6i−2i 2 =8+4i Divide by 4 𝑖 4i: 8 + 4 𝑖 4 𝑖 = 2 − 𝑖 4i 8+4i =2−i