Question 1 Explain how to multiply complex numbers and give an example. Answer 1 To multiply two complex numbers, we use the distributive property. Multiply the real and imaginary parts of the first complex number by the real and imaginary parts of the second complex number, and simplify using 𝑖 2 = − 1 i 2 =−1. Example: Let 𝑧 1 = 2 + 3 𝑖 z 1 =2+3i and 𝑧 2 = 1 + 2 𝑖 z 2 =1+2i. 𝑧 1 ⋅ 𝑧 2 = ( 2 + 3 𝑖 ) ( 1 + 2 𝑖 ) = 2 ⋅ 1 + 2 ⋅ 2 𝑖 + 3 𝑖 ⋅ 1 + 3 𝑖 ⋅ 2 𝑖 = 2 + 4 𝑖 + 3 𝑖 + 6 𝑖 2 z 1 ⋅z 2 =(2+3i)(1+2i)=2⋅1+2⋅2i+3i⋅1+3i⋅2i=2+4i+3i+6i 2 Since 𝑖 2 = − 1 i 2 =−1, = 2 + 4 𝑖 + 3 𝑖 − 6 = − 4 + 7 𝑖 =2+4i+3i−6=−4+7i So, 𝑧 1 ⋅ 𝑧 2 = − 4 + 7 𝑖 z 1 ⋅z 2 =−4+7i. Question 2 To divide complex numbers, multiply the numerator and denominator by the complex ________ of the ________. Answer 1 To divide complex numbers, multiply the numerator and denominator by the complex conjugate of the denominator. Question 3 Graphing Complex Numbers in Exercises 65–68, sketch the graph of all complex numbers 𝑧 z satisfying the given condition. Answer 1 (65) ∣ 𝑧 ∣ = 2 ∣z∣=2 Sketch a circle of radius 2 centered at the origin. (66) ∣ 𝑧 ∣ = 3 ∣z∣=3 Sketch a circle of radius 3 centered at the origin. (67) arg ⁡ ( 𝑧 ) = 𝜋 6 arg(z)= 6 π Draw a line through the origin that makes an angle of 𝜋 6 6 π with the positive real axis. Question 4 Write the complex number in trigonometric form 𝑟 ( cos ⁡ 𝜃 + 𝑖 sin ⁡ 𝜃 ) r(cosθ+isinθ), with 𝜃 θ in the interval [ 0 ∘ , 36 0 ∘ ] [0 ∘ ,360 ∘ ]: 5 3 + 5 𝑖 5 3 +5i. Answer 1 Find the modulus 𝑟 r: 𝑟 = ( 5 3 ) 2 + 5 2 = 75 + 25 = 100 = 10 r= (5 3 ) 2 +5 2 = 75+25 = 100 =10 Find the argument 𝜃 θ: 𝜃 = tan ⁡ − 1 ( 5 5 3 ) = tan ⁡ − 1 ( 1 3 ) = 3 0 ∘ θ=tan −1 ( 5 3 5 )=tan −1 ( 3 1 )=30 ∘ So, the trigonometric form is 10 ( cos ⁡ 3 0 ∘ + 𝑖 sin ⁡ 3 0 ∘ ) 10(cos30 ∘ +isin30 ∘ ). Question 5 Use De Moivre's Theorem to find the indicated power of the complex number. Write the answer in rectangular form: ( 3 − 𝑖 ) 6 ( 3 −i) 6 . Answer 1 Convert 3 − 𝑖 3 −i to polar form: Modulus: 𝑟 = ( 3 ) 2 + ( − 1 ) 2 = 3 + 1 = 2 r= ( 3 ) 2 +(−1) 2 = 3+1 =2 Argument: 𝜃 = tan ⁡ − 1 ( − 1 3 ) = − 𝜋 6 θ=tan −1 ( 3 −1 )=− 6 π Using De Moivre's Theorem: ( 3 − 𝑖 ) 6 = ( 2 ) 6 ( cos ⁡ ( 6 ⋅ − 𝜋 6 ) + 𝑖 sin ⁡ ( 6 ⋅ − 𝜋 6 ) ) ( 3 −i) 6 =(2) 6 (cos(6⋅− 6 π )+isin(6⋅− 6 π )) = 64 ( cos ⁡ ( − 𝜋 ) + 𝑖 sin ⁡ ( − 𝜋 ) ) = 64 ( − 1 + 0 𝑖 ) = − 64 =64(cos(−π)+isin(−π))=64(−1+0i)=−64 So, the answer is − 64 −64