Questions and Answers #8 Complex Numbers

Questions and Answers Sheet 8 Complex Numbers Question #1 Plot the complex number. Then write the trigonometric form of the complex number. 5 - 5i Answer: Question #2 Plot the complex number. Then write the trigonometric form of the complex number. 1 − √3𝑖 Answer: Question #3 Calculate and write in cartesian form a) (2 + 𝑖)(1 − 2𝑖) b) 3+𝑖 4−3𝑖 Answer: a) The complex numbers in cartesian form are 𝑧1 = 2 + 𝑖 𝑧2 = 1 − 2𝑖 𝑧1 × 𝑧2 = (2 + 𝑖)(1 − 2𝑖) = (2)(1) + (𝑖)(1) + (2)(−2𝑖) + (𝑖)(−2𝑖) = 2 + 𝑖 − 4𝑖 − 2𝑖 2 = 2 − 3𝑖 − 2(−1) ⋯ 𝑖 2 = −1 = 2 − 3𝑖 + 2 = 4 − 3𝑖 Hence, (2 + 𝑖)(1 − 2𝑖) = 4 − 3𝑖 Question #4 Calculating probabilities for complex random variables I am having some trouble understanding/formulating how one computes probabilites given a (somehow complex) continuous random variable. For example, if I define a random variable Z as: 𝑌 = 10(2 + 𝜇 + 𝜎𝑋) where X has standard normal distribution. How would I, for example, formulate/define P(Y>0) for some 𝜇 and 𝜎? Answer: Assuming 𝜎 > 0 , from 𝑌 = 10(2 + 𝜇 + 𝜎𝑋) you could say 𝑋 = So 𝑃(𝑌 > 0) is equivalent to 𝑃 (𝑋 > −2−𝜇 𝜎 ) = 1−Φ( −2−𝜇 𝜎 𝑌−20−10𝜇 10𝜎 ) Question #5 Sums and Differences. Evaluate the sum or difference, and write the result in the form a + bi . (3 + 2𝑖) + 5𝑖 and 3𝑖 − (2 − 3𝑖) Answer: We have to write the complex numbers in the form of a+bi where sums or differences are given as: (3+2i)+5i and 3i-(2-3i) We know the addition or subtraction of complex numbers are done through corresponding terms that is real part with real part and imaginary part with imaginary part. Example: 𝑥 + 𝑦𝑖 + 𝑧 + 𝑤𝑖 = (𝑥 + 𝑧) + (𝑦 + 𝑤)𝑖 So solving first part, (3 + 2𝑖) + 5𝑖 = 3 + 2𝑖 + 5𝑖 = 3 + (2 + 5)𝑖 = 3 + 7𝑖 Hence, answer in form of a+bi i 3+7i . Solving second part, 3𝑖 − (2 − 3𝑖) = 3𝑖 − 2 + 3𝑖 = −2 + (3 + 3)𝑖 = −2 + 6𝑖 Hence, answer in form of a+bi is -2+6i . Question #6 Find zw and 𝑧 = 7 (cos 𝜋 4 𝑧 𝑤 . Write each answer in polar form and in exponential form. + 𝑖 sin 𝜋 4 ) 𝑤 = 7 (cos 𝜋 10 + 𝑖 sin 𝜋 10 ) The product zw in polar form is ? and in exponential form is ?. Answer: Use product of complex numbers in polar form 𝑧𝑤 = 𝑟1𝑟2[cos(𝜃1 + 𝜃2) + 𝑖 sin(𝜃1 + 𝜃2)] Use product of complex numbers in polar form z r1 = [cos( 1 −  2) + i sin( 1 −  2) w r2 z r1 = [cos( 1 −  2) + i sin( 1 −  2) w r2 in exponential form 𝑧 = 𝑎𝑒 𝑖𝜃 𝑤 = 𝑏𝑒 𝑖𝜙 𝜋 𝜋 4 10 𝑧𝑤 = 7 ⋅ 7 [cos ( + 𝑧𝑤 = 49 [cos 𝑧𝑤 = 49 [cos 𝜋 + 𝑖 sin 2 7𝜋 20 + 𝑖 sin 𝜋 𝑤 = cos 3𝜋 20 + 𝑖 sin 𝜋 4 10 )] ] 2 7𝜋 20 Similarly we will solve for 𝑧 𝜋 ) + 𝑖 sin ( + 3𝜋 ] 𝑧 𝑤 20 now we will find exponential form which will be 7𝜋 𝑧𝑤 = 49 ⋅ 𝑒 𝑖 ⋅ ( ) 20 Question #7 Find the distance between the complex numbers in the complex plane. 1 + 4i, -1 + 7i Answer: The distance between the two complex numbers, 𝑧1 = 𝑎 + 𝑏𝑖 and 𝑧2 = 𝑐 + 𝑑𝑖 in the complex plane is 𝐷 = √(𝑐 − 𝑎)2 + (𝑑 − 𝑏)2 In the cartesian plane, the distance between one point and another is (𝑥1 , 𝑦1 ), and (𝑥2 , 𝑦2 ) is the distance formula is 𝐷 = √(𝑥2 − 𝑥1 )2 + (𝑦2 − 𝑦1 )2 To find the distance between the two complex number using the formula 𝐷 = √(𝑐 − 𝑎)2 + (𝑑 − 𝑏)2 where 𝑧1 = 1 + 4𝑖 and 𝑧2 = −1 + 7𝑖 substitute a=1, b=4, c=-1, and d=7 in the distance formula a=1 b=4 c=-1 d=7 𝐷 = √(𝑐 − 𝑎)2 + (𝑑 − 𝑏)2 = √(−1 − 1)2 + (7 − 4)2 = √−22 + 32 = √4 + 9 Answer: 𝐷 = √13 Question #8 What is the magnitude of two complex numbers multiplied together, 𝑍1 ⋅ 𝑍2 ? 𝑍1 = 6.77 + 𝑗 ⋅ 0.09, 𝑍2 = −3.27 + 𝑗 ⋅ 6.98 Three significant digits, fixed point notation. Answer: 𝑍1 = 6.77 + 𝑗 × 0.09 𝑍2 = −3.27 + 6.98𝑗 |𝑍1 | = √(6.77)2 + (0.09)2 |𝑍1 | = √45.841 |𝑍1 | = 6.77 |𝑍1 | = 6.77 |𝑍2 | = √(−3.27)2 + (6.98)2 |𝑍2 | = √59.4 |𝑍2 | = 7.71 |𝑍1 𝑍2 | = |𝑍1 ||𝑍2 | |𝑍1 𝑍2 | = 6.77 × 7.71 |𝑍1 𝑍2 | = 52.2 Question #9 Add (plus) the following complex numbers: a) 9+j3 and 5+j8 b) -18+j23 and 30-j15 c) 2 < 45∘ and 20 < 32∘ Answer: we have to add the complex numbers a) (9 + 𝑗3) + (5 + 𝑗8) = 9 + 5 + 𝑗3 + 𝑗8 = 14 + 𝑗11 b) (-18+j23)+(30-j15) =-18+30+j23-j15 =12+j8 c) (2, ∠45∘ ) + (20, ∠32∘ ) = 2(cos 4 5∘ + 𝑗 sin 4 5∘ ) + 20(cos 3 2∘ + 𝑗 sin 3 2∘ ) = √2 + 𝑗√2 + 16.96 + 𝑗10.6 = 1.41 + 𝑗1.41 + 16.96 + 𝑗10.6 = 18.37 + 𝑗12.01 Question #10 Subtract the two complex numbers: (3 + 6i) - (5i) . What do you notice? Answer: When we perform operations on complex numbers, we perform it independently. For example, 𝐴 = (𝑖 + 𝑗) + (2𝑖 + 3𝑗) 𝐴 = (3𝑖 + 4𝑗) Here, we have added 1i to 2i and 1j to 3j and the final complex number is A = (3i + 4j) Let's assume, A = (3 + 6i) B = (5i) now, A - B = C C = (3 + 6i) - (0 + 5i) C = 3 - 0 + (6 - 5)i Answer: C = 3 + 1i Question #11 Find the product of the complex numbers 𝑧1 = cos 𝜋 4 + 𝑖 sin 𝜋 4 and 𝑧2 = cos 𝜋 3 + 𝑖 sin 𝜋 3 . Leave answers in polar form. Answer: To find the product of the complex numbers 𝜋 𝜋 𝑧1 = cos + 𝑖 sin 4 4 𝜋 𝜋 𝑧2 = cos + 𝑖 sin 4 3 using identities cos 𝐴 cos 𝐵 − sin 𝐴 sin 𝐵 = cos(𝐴 + 𝐵) cos 𝐴 sin 𝐵 + sin 𝐴 cos 𝐵 = sin(𝐴 + 𝐵) 𝜋 𝜋 𝜋 𝜋 𝑧1 × 𝑧2 = cos ( + ) + 𝑖 sin ( + ) 4 3 4 3 7𝜋 7𝜋 𝑧1 × 𝑧2 = cos + 𝑖 sin 12 12 Question #12 Find the distance between the complex numbers in the complex plane. -8 + i, -3 + 8i Answer: The real parts of both complex numbers are -8 and -3 and the imaginary parts are 1 and 8. Now the distance is, 2 𝑑 = √(−3 − (−8)) + (8 − 1)2 = √25 + 49 = √74 Question #13 Solve the equation 𝑥 2 + 12 = 0 ? Answer: We have: 𝑥 2 = −12 Taking square roots we have: 𝑥 = ±√−12 = ±√3 × 4𝑖 = ±2√3𝑖 Question #14 Find the roots of 𝑧 5 + 1 = 0 ? Answer: Let 𝜔 = −1 , and then 𝑧 5 = 𝜔 And we will put the complex number into polar form (visually): |𝜔| = 1 arg(𝜔) = So then in polar form we have: 𝜔 = cos(𝜋) + 𝑖 sin(𝜋) We now want to solve the equation 𝑧 5 = 𝜔 for z (to gain 5 solutions): 𝑧 5 = cos(𝜋) + 𝑖 sin(𝜋) Whenever dealing with complex variable equation such as this it is essential to remember that the complex exponential (and therefore the polar representation) has a period of 2𝜋 , so we can equivalently write (incorporating the periodicity): 𝑧 5 = cos(𝜋 + 2𝑛𝜋) + 𝑖 sin(𝜋 + 2𝑛𝜋) 𝑛 ∈ 𝑍 By De Moivre's Theorem we can write this as: 1 𝑧 = (cos(𝜋 + 2𝑛𝜋) + 𝑖 sin(𝜋 + 2𝑛𝜋))5 = cos ( 𝜋+2𝑛𝜋 5 ) + 𝑖 sin ( 𝜋+2𝑛𝜋 5 ) = cos 𝜃 + 𝑖𝑠𝑖𝑛 𝜃 where 𝜃 = (2𝑛+1)𝜋 Working to 3dp we get: Put: 𝑛 = −2 ⇒ 𝜃 = − ∴ 𝑧 = cos (− 3𝜋 5 3𝜋 3𝜋 ) + 𝑖 sin (− ) 5 5 ∴ 𝑧 = −0.309 − 0.951𝑖 𝑛 = −1 ⇒ 𝜃 = − 𝜋 5 𝜋 𝜋 5 5 ∴ 𝑧 = cos (− ) + 𝑖 sin (− ) ∴ 𝑧 = 0.809 − 0.588𝑖 𝑛=0⇒𝜃= 𝜋 5 𝑝𝑖 𝜋 5 5 ∴ 𝑧 = cos ( ) + 𝑖 sin ( ) 5 ∴ 𝑧 = 0.809 + 0.588𝑖 𝑛=1⇒𝜃= 3𝜋 5 3𝜋 3𝜋 5 5 ∴ 𝑧 = cos ( ) + 𝑖 sin ( ) ∴ 𝑧 = −0.309 + 0.951 𝑛=2⇒𝜃=𝜋 ∴ 𝑧 = cos 𝜋 + 𝑖 𝑠𝑖𝑛 𝜋 \therefore z=-1 After which the pattern continues. Question #15 What roots of 𝑧 4 + 𝑧 2 + 1 = 0 satisfy |𝑧| < 1 ? Answer: Calling 𝑢 = 𝑧 2 we have 𝑢2 + 𝑢 + 1 = 0 solving for u 𝑢= 1 (−1 ± 𝑖√3) = 𝑒 ±𝑖(𝜙+2𝑘𝜋) 2 with 𝜙 = arctan √3 then 𝑖 𝑧 2 = 𝑒 ±𝑖(𝜙+2𝑘𝜋) so 𝑧 = 𝑒 ±2 (𝜙+2𝑘𝜋) but |𝑒 𝑖𝑥 | = 1 so all four roots have unit modulus and no root is in |z|<1 Question #16 How do you find all solutions to 3 +1=0? Answer: Using synthetic division and the fact that x=-1 is obviously a solution we find that we can expand this to: (𝑥 + 1)(𝑥 2 − 𝑥 + 1) = 0 In order to have LHS=RHS need one of the brackets to be equal to zero, ie 1) (𝑥 + 1) = 0 2) (𝑥 2 − 𝑥 + 1) = 0 Fro m1 we note that x=-1 is a solution. We shall solve 2 using the quadratic formula: 𝑥2 − 𝑥 + 1 = 0 𝑥= 1±√(−1)2 −4(1)(1) 2 = 1±√−3 2 = 1±√3𝑖 2 Question #17 If 𝑧 ∈ 𝐶 then what is √𝑧 2 ? Answer: I'm not sure why specifying 𝑧 ∈ 𝐶 is significant. If z=a+bi then 𝑧 2 = 𝑎2 + 2𝑎𝑏𝑖 − 𝑏 2 Any value (𝑖𝑛 𝐶, ) 𝑧̂ for which 𝑧̂2 = 𝑎2 + 2𝑏𝑖 − 𝑏 2 should be a square root of z The two possible 𝑧̂ values are 𝑧̂ = 𝑎 + 𝑏𝑖 = 𝑧 and 𝑧̂ = −𝑎 − 𝑏𝑖 = −𝑧 Question #18 How do you find the absolute value of -2-i? Answer: Absolute value of a complex number a+ib is written as |a+ib| and its value as √𝑎2 + 𝑏 2 Hence, absolute value of -2-i is √(−2)2 + (−1)2 = √4 + 1 = √5 Question #19 How do you find the absolute value of -7+i ? Answer: The absolute value of 𝑧 = 𝑥 + 𝑦𝑖 is |𝑧| = |𝑥 + 𝑦𝑖| = √𝑥 2 + 𝑦 2 ⇒ |−7 + 𝑖| = √(−7)2 + 12 = √50 = 5√2 Question #20 How do you find the absolute value of 2+5i ? Answer: The modulus of a complex number a+bi is given by: |𝑎 + 𝑏𝑖| = √𝑎2 + 𝑏 2 Thus: |2 + 5𝑖| = √22 + 52 = √4 + 25 = √29