Question 1 How do you find the absolute value of the complex number 𝑧 = 3 − 4 𝑖 z=3−4i? Answer 1 Step 1: One way to do this is with the general formula ∣ 𝑎 + 𝑏 𝑖 ∣ = 𝑎 2 + 𝑏 2 ∣a+bi∣= a 2 +b 2 , so ∣ 3 − 4 𝑖 ∣ = 3 2 + ( − 4 ) 2 = 9 + 16 = 25 = 5 ∣3−4i∣= 3 2 +(−4) 2 = 9+16 = 25 =5 Question 2 What are complex numbers? ln ⁡ ( 1 − 𝑖 ) ln(1−i) Answer 1 Complex numbers are numbers that can be expressed in the form 𝑎 + 𝑏 𝑖 a+bi, where 𝑎 a and 𝑏 b are real numbers, and 𝑖 i is the imaginary unit with the property 𝑖 2 = − 1 i 2 =−1. Answer 2 To find ln ⁡ ( 1 − 𝑖 ) ln(1−i): Rewrite 1 − 𝑖 1−i in polar form: The modulus 𝑟 = ∣ 1 − 𝑖 ∣ = 1 2 + ( − 1 ) 2 = 2 r=∣1−i∣= 1 2 +(−1) 2 = 2 . The argument 𝜃 = tan ⁡ − 1 ( − 1 1 ) = − 𝜋 4 θ=tan −1 ( 1 −1 )=− 4 π . Then, ln ⁡ ( 1 − 𝑖 ) = ln ⁡ ( 2 ) + 𝑖 ( − 𝜋 4 + 2 𝑘 𝜋 ) = ln ⁡ 2 2 − 𝜋 4 𝑖 + 2 𝑘 𝜋 𝑖 , 𝑘 ∈ 𝑍 . ln(1−i)=ln( 2 )+i(− 4 π +2kπ)= 2 ln2 − 4 π i+2kπi,k∈Z. Question 3 Why does 𝑖 2 = − 1 i 2 =−1? Answer 1 This question is similar to asking "why does 3 2 = 9 3 2 =9?" The definition of 𝑖 i is that it is a number which satisfies 𝑖 2 = − 1 i 2 =−1. This definition was chosen to enable solutions for equations like 𝑥 2 + 1 = 0 x 2 +1=0, which have no solutions among real numbers. By defining 𝑖 2 = − 1 i 2 =−1, mathematicians expanded the number system to include solutions for such equations, creating the complex numbers. Question 4 Multiply each complex number by its complex conjugate. a) 3 + 6 𝑖 3+6i b) 2 − 5 𝑖 2−5i Answer 1 Part a Given 3 + 6 𝑖 3+6i, its conjugate is 3 − 6 𝑖 3−6i. ( 3 + 6 𝑖 ) ( 3 − 6 𝑖 ) = 3 2 − ( 6 𝑖 ) 2 = 9 + 36 = 45 (3+6i)(3−6i)=3 2 −(6i) 2 =9+36=45 Part b Given 2 − 5 𝑖 2−5i, its conjugate is 2 + 5 𝑖 2+5i. ( 2 − 5 𝑖 ) ( 2 + 5 𝑖 ) = 2 2 − ( 5 𝑖 ) 2 = 4 + 25 = 29 (2−5i)(2+5i)=2 2 −(5i) 2 =4+25=29 Question 5 How do you express the complex number in rectangular form: 𝑧 = 5 ( cos ⁡ 12 0 ∘ + 𝑖 sin ⁡ 12 0 ∘ ) z=5(cos120 ∘ +isin120 ∘ )? Answer 1 To convert from polar to rectangular form: 𝑥 = 𝑟 cos ⁡ 𝜃 , 𝑦 = 𝑟 sin ⁡ 𝜃 x=rcosθ,y=rsinθ Here, 𝑟 = 5 r=5 and 𝜃 = 12 0 ∘ θ=120 ∘ . 𝑥 = 5 cos ⁡ 12 0 ∘ = 5 × − 1 2 = − 5 2 x=5cos120 ∘ =5×− 2 1 =− 2 5 𝑦 = 5 sin ⁡ 12 0 ∘ = 5 × 3 2 = 5 3 2 y=5sin120 ∘ =5× 2 3 = 2 5 3 Thus, 𝑧 = − 5 2 + 5 3 2 𝑖 z=− 2 5 + 2 5 3 i