Questions and Answers #2 Complex Variables

Questions and Answers Sheet 2 Complex Variables Question #1 Find all the complex roots. Leave your answers in polar form with the argument in degrees. The complex fourth roots of 100 + 100√3𝑖 Answer: Step 1 Consider the given complex number 100 + 100√3𝑖 Step 2 Convert the complex number into polar form. We have that 𝑎 = 100 𝑎𝑛𝑑 𝑏 = 100√3 2 𝑟 = √𝑎2 + 𝑏 2 = √(100)2 + (100√3) = 200 𝜃 = tan−1 ( 100√3 100 )= 𝜋 3 = 60∘ ∴ 100 + 100√3𝑖 = 200(cos 6 0∘ + 𝑖 sin 6 0∘ ) Step 3 Find the fourth roots of complex number. The nth roots of a complex number is given by 𝑛 𝑧𝑘 = √𝑟 [cos ( 𝜃+360∘ 𝑘 𝑛 ) + 𝑖 sin ( 𝜃+360∘ 𝑘 We have 𝑛 = 4, 𝑟 = 200, 𝜃 = 60 4 𝑧𝑘 = √200 [cos ( 60∘ +360∘ 𝑘 4 𝑛 )] , 𝑘 = 0,1, ⋯ , 𝑛 − 1 ∘ ) + 𝑖 sin ( 60∘ +360∘ 𝑘 4 )] , 𝑘 = 0,1,2,3 Therefore, 3 𝑧𝑘 = 24 ⋅ √5[cos(15∘ + 90∘ 𝑘) + 𝑖 sin(15∘ + 90∘ 𝑘)], 𝑘 = 0,1,2,3 Question #2 Which of the following is true? 𝑑𝑦 𝑑𝑡 𝑑𝑦 𝑑𝑡 𝑑𝑦 𝑑𝑡 𝑑𝑦 𝑑𝑡 𝑑𝑦 𝑑𝑡 + = = 𝑦 1−𝑡 𝑡 = 3𝑡 + 4 is best solved with undetermined coefficients. 𝑦−𝑦𝑡 2 𝑡 𝑦 is best solved with separation of variables. is best solved with integrating factors. = 2𝑦 − sin(5𝑡) is best solved with separation of variables. = 𝑡 𝑦−𝑦𝑡 2 is best solved with undetermined coefficients. Answer: Step 1 We have to give which is true 𝑑𝑦 𝑑𝑡 𝑑𝑦 𝑑𝑡 𝑑𝑦 𝑑𝑡 + = = 𝑦 1−𝑡 𝑡 = 3𝑡 + 4 is best solved with undetermined coefficients. 𝑦−𝑦𝑡 2 𝑡 𝑦 is best solved with separation of variables. is best solved with integrating factors. 𝑑𝑦 𝑑𝑡 𝑑𝑦 𝑑𝑡 = 2𝑦 − sin(5𝑡) is best solved with separation of variables. = 𝑡 𝑦−𝑦𝑡 2 is best solved with undetermined coefficients. Step 2 Answer=(2) (2) 𝑑𝑦 𝑑𝑡 = 𝑡 𝑦−𝑦𝑡 2 𝑡 ⇒ 𝑦𝑑𝑦 = ( 1−𝑡 2 = 𝑡 𝑦(1−𝑡 2 ) ) 𝑑𝑡(True) This is best solved with separation of variables Step 3 Since , (1) is best solved with Integrating factor (3) is best solved with separation of variables (4) is best solved with integrating factor (5) is best solved with separation of variables Question #3 Find all complex zeros of the polynomial function. Give exact values. List multiple zeros as necessary. 𝑓(𝑥) = 𝑥 4 − 5𝑥 3 − 36𝑥 2 + 272𝑥 − 448 All complex zeros are=? Answer: Step 1 Given, 𝑓(𝑥) = 𝑥 4 − 5𝑥 3 − 36𝑥 2 + 272𝑥 − 448 Factorizing f(x) as follows: 𝑥 4 − 5𝑥 3 − 36𝑥 2 + 272𝑥 − 448 = 𝑥 3 (𝑥 − 4) − 𝑥 2 (𝑥 − 4) − 40𝑥(𝑥 − 4) + 112(𝑥 − 4) = (𝑥 − 4)(𝑥 3 − 𝑥 2 − 40𝑥 + 112) = (𝑥 − 4)(𝑥 − 4)(𝑥 2 + 3𝑥 − 28) = (𝑥 − 4)(𝑥 − 4)(𝑥 − 4)(𝑥 + 7)(𝑥 − 4) = (𝑥 − 4)3 (𝑥 + 7) Step 2 Equating it to zero (𝑥 − 4)3 (𝑥 + 7) = 0 x-4=0 x+7=0 x=4; x=-7 Thus, there are no complex zeros. Question #4 Lynbrook West, an apartment complex , has 100 two-bedroom units. The montly profit (in dollars) realized from renting out x apartments is given by the following function. 𝑃(𝑥) = −12𝑥 2 + 2136𝑥 − 41000 To maximize the monthly rental profit , how many units should be rented out? What is the maximum monthly profit realizable? Answer: Given profit function, 𝑃(𝑥) = −12𝑥 2 + 2136𝑥 − 41000 To maximize profit 𝑑 𝑑𝑥 ⇒ 𝑃(𝑥) = 0 𝑑 𝑑𝑥 (−12𝑥 2 + 2136𝑥 − 41000) = 0 ⇒ −24𝑥 + 2136 = 0 ⇒ −24𝑥 = −2136 −2136 ⇒𝑥= −24 x=89 Here, 𝑑 2 𝑃(𝑥) 𝑑𝑥 2 = −24 < 0 as per second derirative test profit is maximum at x=89 Also , maximum profit = 𝑃(89) = −12(89)2 + 2136(89) − 41000 =54052 To maximize the montly rental profit, 89 units should be rented out, 54052$ be the maximum monthly profit Question #5 Lynbrook West, an apartment complex, has 100 two-bedroom units. The monthly profit (in dollars) realized from renting x apartments is represented by the following function. 𝑃(𝑥) = −11𝑥 2 + 1830𝑥 − 34000 (a) What is the actual profit realized from renting the 41st unit, assuming that 40 units have already been rented? Answer: Step 1 Consider the given equation for the monthly benefit by renting x apartments: 𝑃(𝑥) = −11𝑥 2 + 1830𝑥 − 34000 Step 2 Now calculate the profit by renting 40 apartments: 𝑃(40) = −11 ⋅ 402 + 1830 ⋅ 40 − 34000 = −17600 + 73200 − 34000 = 21600 Step 3 Also calculate the profit by renting 41 apartments: 𝑃(41) = −11 ⋅ 412 + 1830 ⋅ 41 − 34000 = −18497 + 75030 − 34000 = 22539 Step 4 Therefore,actual benefit realized by renting 41st apartment is calculate as follows: Profit = 𝑃(41) − 𝑃(40) = 22539 − 21600 = 939 Question #6 For each of the systems of equations that follow, use Gaussian elimination to obtain an equivalent system whose coefficient matrix is in row echelon form. Indicate whether the system is consistent. If the system is consistent and involves no free variables, use back substitution to find the unique solution. If the system is consistent and there are free variables, transform it to reduced row echelon form and find all solutions. 2𝑥1 − 3𝑥2 = 5 −4𝑥1 + 6𝑥2 = 8 Answer: Given: The given system of equations is 2𝑥1 − 3𝑥2 = 5 −4𝑥1 + 6𝑥2 = 8 Theorem used: Test for Consistency: The following conditions are equivalent: (a) The matrix equation Ax=b is consistent (b) The vector b is a linear combination of column of A. (c) The reduced row echelon form of the augmented matrix [A b] has no row of the form [00 ⋯ 0𝑑], where 𝑑 ≠0 Calculation: Write the augmented matrix as follows Apply the Gaussian elimination algorithm to transform the matrix into one in the row echelon form. So,this matrix is the reduced row echelon form of the augmented matrix of given system. It is observed that the augmented matrix contains a row in which the only nonzero entry appears in the last column. Therefore, the given system of equations must be inconsistent. So, there is no solution. Question #7 Complex numbers can serve as entries in a matrix just as well as real numbers.Compute the expressions in Problems 51-53 , where 2i  1 + i  1 −i  A= and B =     2 2 − 3i  2i 1 + i  51.A+2B 52.AB 53. BA Answer: Step 1 Given: 2i  1 + i  1 −i  A= and B =     2 2 − 3i  2i 1 + i  Step 2 2i  1 + i  1 −i  A= and B =     2 2 − 3i  2i 1 + i  51.A+2B 2i  2i   2 −2i  1 + i + 2 2i − 2i   3 + i 0  1 + i  1 −i  1 + i + 2 = + = =  2 2 − 3i  2i 1 + i   2 2 − 3i  4i 2 + 2i   2 + 4i 2 − 3i + 2 + 2i  2 + 4i 4 − i              Step 3 2i   1 −i   (1 + i)(1) + 2i  2i (1 + i) + 2i(1 + i)  1 + i AB =  =      2 2 − 3i   2i 1 + i   2 1 + (2 − 3i)  2i 2  (−i) + (2 − 3i)(1 + i)  i −3 i(1 + i)   =  2 + 2i(2 − 3i) −2i + (2 − 3i)(1 + i)  52. 2i   1 −i  1 + i BA =      2i 1 + i   2 2 − 3i  1(2i) + ( −i)(2 − 3i)   1(1 + i ) + (−i)(2) 53. =    2i (1 + i ) + (1 + i )  2 2i  2i + (1 + i )(2 − 3i)  1− i −i (2 − 3i) + 2i   =  2 2  2i + 2i + 2 + 2i 4i + (1 + i )(2 − 3i )  Question #8 Use the following linear regression equation to answer the questions. 𝑥1 = 1.3 + 3.0𝑥2 − 8.3𝑥3 + 1.6𝑥4 (a) Which variable is the response variable? (b) Which number is the constant term? List the coefficients with their corresponding explanatory variables. (c) If 𝑥2 = 8, 𝑥3 = 2, and 𝑥4 = 6, what is the predicted value for 𝑥1 ? (Use 1 decimal place) (d) Explain how each coefficient can be thought of as a ''slope'' under certain conditions. Answer: (a) Given, Linear degression equation is 𝑥1 = 1 ⋅ 3 + 3 ⋅ 0𝑥2 − 8 ⋅ 3𝑥3 + 1 ⋅ 6𝑥4 a) The response variable is 𝑥1 and Explanatory variable is 𝑥2 , 𝑥3 , 𝑥4 b) Constant term = 1 ⋅ 3 𝑥2 coefficient = 3 ⋅ 0 𝑥3 coefficient = −8 ⋅ 3 𝑥4 coefficient = 1 ⋅ 6 c) If 𝑥2 = 8, 𝑥3 = 2 𝑎𝑛𝑑 𝑥4 = 6 then 𝑥1 = 1 ⋅ 3 + (3 ⋅ 0 ∗ 8) − (8 ⋅ 3 ∗ 2) + (1 ⋅ 6 ∗ 6) 𝑥1 = 1 ⋅ 3 + 24 − 16 ⋅ 6 + 9 ⋅ 6 𝑥1 = 18 ⋅ 3 d) If we hold all other explanatory variables as fixed constants, then we can look at one coefficient as a ''slope'' Question #9 Second derivatives For the following sets of variables, find all the relevant second derivatives. In all cases, first find general expressions for the second derivatives and then substitute variables at the last step. 𝑠 𝑓(𝑥, 𝑦) = 𝑥 2 𝑦 − 𝑥𝑦 2 ,where 𝑥 = 𝑠𝑡 and 𝑦 = 𝑡 Answer: Step 1 Given function is , 𝑓(𝑥, 𝑦) = 𝑥 2 𝑦 − 𝑥𝑦 2 ,where 𝑥 = 𝑠𝑡 and 𝑦 = The all relevant second derivatives are 𝑓𝑥𝑥 , 𝑓𝑦𝑦 , 𝑓𝑥𝑦 , 𝑓𝑦𝑥 𝑠 𝑡 Step 2 Find partial derivative with respect to x for the function 𝑓(𝑥, 𝑦) = 𝑥 2 𝑦 − 𝑥𝑦 2 ⇒ 𝑓𝑥 = 2𝑥𝑦 − 𝑦 2 Find partial derivative with respect to y for the function 𝑓(𝑥, 𝑦) = 𝑥 2 𝑦 − 𝑥𝑦 2 ⇒ 𝑓 − 𝑦 = 𝑥 2 − 2𝑥𝑦 Differentiate function 𝑓𝑥 = 2𝑥𝑦 − 𝑦 2 with respect to x to get 𝑓𝑥𝑥 ⇒ 𝑓𝑥𝑥 = 2𝑦 Substitute 𝑦 = 𝑠𝑡 in 𝑓𝑥𝑥 = 2𝑦 2𝑠 ⇒ 𝑓𝑥𝑥 = 𝑡 Differentiate function 𝑓𝑦 = 𝑥 2 − 2𝑥𝑦 with respect to y to get 𝑓𝑦𝑦 ⇒ 𝑓𝑦𝑦 = −2𝑥 Substitute 𝑥 = 𝑠𝑡 in 𝑓𝑦𝑦 = −2𝑥 ⇒ 𝑓𝑦𝑦 = −2𝑠𝑡 Differentiate function 𝑓𝑦 = 𝑥 2 − 2𝑥𝑦 with respect to x to get 𝑓𝑥𝑦 ⇒ 𝑓𝑥𝑦 = 2𝑥 − 2𝑦 Substitute 𝑥 = 𝑠𝑡, 𝑦 = ⇒ 𝑓𝑥𝑦 = 2𝑠𝑡 − 2𝑠 𝑠 𝑡 in 𝑓𝑥𝑦 = 2𝑥 − 2𝑦 𝑡 Differentiate function 𝑓𝑥 = 2𝑥𝑦 − 𝑦 2 with respect to y to get 𝑓𝑦𝑥 ⇒ 𝑓𝑦𝑥 = 2𝑥 − 2𝑦 Substitute 𝑥 = 𝑠𝑡, 𝑦 = ⇒ 𝑓𝑦𝑥 = 2𝑠𝑡 − 𝑠 𝑡 in 𝑓𝑦𝑥 = 2𝑥 − 2𝑦 2𝑠 𝑡 Step 3 Therefore, all relevant second derivatives of function 𝑓(𝑥, 𝑦) = 𝑥 2 𝑦 − 𝑥𝑦 2 , where 𝑥 = 𝑠𝑡, 𝑦 2𝑠 2𝑠 2𝑠 𝑓𝑥𝑥 = , 𝑓𝑦𝑦 = −2𝑠𝑡, 𝑓𝑥𝑦 = − 2𝑠𝑡, 𝑓𝑦𝑥 = − 2𝑠𝑡 𝑡 𝑡 𝑡 {𝑠] 𝑡 are, Question #10 4.3-16. Two random variables X and Y have a joint density 10 𝑓𝑥,𝑦 (𝑥, 𝑦) = [𝑢(𝑥) − 𝑢(𝑥 − 4)]𝑢(𝑦)𝑦 3 𝑒𝑥𝑝[−(𝑥 + 1)𝑦 2 ] 4 Find the marginal densities and distributions of X and Y. Answer: Step 1 It is given that: 10 𝑓𝑥𝑦 (𝑥, 𝑦) = [𝑢(𝑥) − 𝑢(𝑥 − 4)]𝑢(𝑦)𝑦 3 𝑒𝑥𝑝[−(𝑥 + 1)𝑦 2 ] 4 The marginal pdf of X is given by: 𝑓𝑥 (𝑋) = ∫𝑦 ∫𝑥𝑦(𝑥, 𝑦)𝑑𝑦 = = 10 4 10 4 [𝑢(𝑥) − (𝑥 − 4)] ∫ 𝑢(𝑦)𝑦 3 𝑒𝑥𝑝[−(𝑥 + 1)𝑦 2 ]𝑑𝑦 [𝑢(𝑥) − 𝑢(𝑥 − 4)]𝑒 −(𝑥+1) ∫ 𝑢 (𝑦)𝑦 3 𝑒 𝑦2 𝑑𝑦 Since the explicit form of the function and its range is not given, the integral cannot be solved further. Step 2 The marginal density of Y is given by: 𝑓𝑦 (𝑌) = ∫𝑥 𝑓𝑥𝑦 (𝑥, 𝑦)𝑑𝑥 = 10 4 10 𝑢(𝑦)𝑦 3 ∫ 𝑒𝑥𝑝 [−(𝑥 + 1)𝑦 2 ][𝑢(𝑥) − 𝑢(𝑥 − 4)]𝑑𝑥 2 𝑢(𝑦)𝑦 3 𝑒 𝑦 ∫[𝑢(𝑥) − 𝑢(𝑥 − 4)]𝑒 −(𝑥+1) 𝑑𝑥 Since the explicit form of the function and its range is not given, the integral cannot be solved further. = 4 Step 3 Hence the marginal densities are: 2 10 𝑓𝑥 (𝑋) = 𝑢(𝑥) − 𝑢(𝑥 − 4)𝑒 −(𝑥+1) ∫ 𝑢(𝑦)𝑦 3 𝑒 𝑦 and 𝑓𝑦 (𝑌) = 4 10 4 2 𝑢(𝑦)𝑦 3 𝑒 𝑦 ∫[𝑢(𝑥) − 𝑢(𝑥 − 4)]𝑒 −(𝑥+1) 𝑑𝑥 Question #11 First determine whether the solutions of the quadratic equation real or complex without solving the equation. Then solve the equation. 3𝑥 2 − 4𝑥 − 15 = 0 Determine whether the solutions of the quaratic equation are real or complex without solving the equation. Answer: Given: 3𝑥 2 − 4𝑥 − 15 = 0 to check whether the solution of the quadratic equation real or complex , we have to find discriminant , D: 𝐷 = 𝑏 2 − 4𝑎𝑐 If 𝐷 < 0 ⇒ complex roots If 𝐷 ≥ 0 ⇒ Real root 3𝑥 2 − 4𝑥 − 15 = 0 a=3 , b=-4 , c=-15 ∴ 𝐷 = 𝑏 2 − 4𝑎𝑐 = (−4)2 − 4(3) ⋅ (−15) =16+180 =196 > 0 Hence roots are real because discriminant is equal to 196 ∴𝑥= = = −𝑏±√𝑏 2 −4𝑎𝑐 2𝑎 −(−4)±√196 2×3 4±14 6 18 −10 = or 6 6 −5 =3 or 3 ∴ Roots are 3, −5 3 Question #12 Convert the equalities below equalities by adding sack variables. Use 𝑠1 and 𝑠2 for your slack variables. 3𝑥1 + 9𝑥2 ≤ 42 converts to () =42 15𝑥1 + 7𝑥2 ≤ 38 converts to () = 38 Answer: Step 1 Solution: Given: 3𝑥1 + 9𝑥2 ≤ 42 ⇒ (1) 15𝑥1 + 7𝑥2 ≤ 38 ⇒ (2) Now to equalize the above equation , we add 𝑠1 and 𝑠2 to the above eqn (1) and (2) respectively, 3𝑥1 + 9𝑥2 + 𝑠1 = 42 ⇒ (1𝑎) 15𝑥1 + 7𝑥2 + 𝑠2 = 38 ⇒ (2𝑓) Step 2 ∴ For initial simplex , basis feasible solution in 𝑥1 = 0 and 𝑥2 = 0 ⇒ 𝑠1 = 42 and 𝑠2 = 38 So, answer. 3𝑥1 + 9𝑥2 + 𝑠1 = 42 15𝑥1 + 7𝑥2 + 𝑠2 = 38 Question #13 Consider the system 𝑥̇ = 𝑥 − 𝑦, 𝑦̇ = 𝑥 + 𝑦 a) Write the system in matrix form and find the eigenvalues and eigenvectors (Note: they will be complex valued) b) Write down the general solution for the system of differential equations using only real valued functions. [Hint : use the equality 𝑒 𝑖𝜔𝑡 = cos(𝜔𝑡) + 𝑖 sin(𝜔𝑡) ] Answer: Step 1 Given: 𝑑𝑥 =𝑥−𝑦 𝑑𝑡 𝑑𝑦 =𝑥+𝑦 𝑑𝑡 To find: (a) = Write system in matrix form , find eigenvalues and eigenvectors. (b). = General solution. Step 2 Solution: a) 𝑑𝑥 𝑑𝑡 𝑑𝑦 𝑑𝑡 =𝑥−𝑦 =𝑥+𝑦 ⇒ 𝑥 ′ (𝑡) = 𝐴𝑥(𝑡), where A is matrix  x  1 −1  x   =    y 1 1   y  Required matrix form of the system. Eigenvalues of matrix A : ⇒ |𝐴 − 𝑥𝐼| = 0  1 − x −1 =0 1 1− x ⇒ (1 − 𝑥)2 + 1 = 0 ⇒ 𝑥 2 − 2𝑥 + 2 = 0 ⇒𝑥= 2±√4−8 2 = 𝑥 = 1 ±𝑖 Hence the eigenvalues are in complex from which are 1+i , 1-i Eigenvector corresponding to 1+i : [𝐴 − (1 + 𝑖)𝐼]𝑣1 = 0 −1   x  1 − 1 − i =0  1 1 − 1 − i   y    −i −1  x     =0  1 −i   y  𝑅2 ⇒ 𝑖𝑅2 + 𝑅1  −i −1  x     =0  0 0   y ⇒ y is free variable , let 𝑦 = 1 ⇒ −𝑖𝑥 − 1 = 0 1 ⇒𝑥= =𝑖 −𝑖 i   v1 =   1 Eigenvector corresponding to 1-i: [𝐴 − (1 − 𝑖)𝐼]𝑣2 = 0 −1   x  1 − 1 + i =0  1 1 − 1 + i   y   i −1  x     =0 1 i   y  𝑅2 ⇒ 𝑖𝑅2 − 𝑅1  i −1  x     =0 0 0   y  ⇒ y is free variable , let 𝑦 = 1 we get ⇒ 𝑖𝑥 − 1 = 0 1 ⇒ 𝑥 = = −𝑖 𝑖  −i   v1 =   1 (b) : Generel solution: i  −i  i   −i  x (t ) = C1e(1+i )t   + C2e(1−i )t   = C1et (cos t + i sin t )   + C2et (cos t − i sin t )   1 1 1 1 Question #14 Solve for i in terms of other variables: (1+𝑖)𝑛 −1 𝐹 = 𝐶[ 𝑖 ] Answer: Step 1 since, we know Binomial Expansion for (1 + 𝑥 𝑛 ) is Replace 𝑛(𝑛 − 1) 2 𝑛(𝑛 − 1)(𝑛 − 2) 3 (1 + 𝑥)𝑛 = 1 + 𝑛𝑥 + 𝑥 + 𝑥 + ⋯ + 𝑥𝑛 2! 3! Replace n by (n-1), we get (𝑛 − 1)(𝑛 − 2) 2 (𝑛 − 1)(𝑛 − 2)(𝑛 − 3) 3 (1 + 𝑥)𝑛−1 = 1 + (𝑛 − 1)𝑥 + 𝑥 + 𝑥 + ⋯ + 𝑥 𝑛−1 2! 3! Step 2 (1+𝑖)𝑛 −1 Now, given that 𝐹 = 𝐶 [ 𝑖 ] Since, we know (1 + 𝑖)𝑛 = 1 + 𝑛𝑖 + 𝑛(𝑛−1) 2 𝑖 2! + ⋯ + 𝑖𝑛 then (1 + 𝑖)𝑛 − 1 = 1 + 𝑛𝑖 + 𝑛(𝑛−1) 2 𝑖 2! + ⋯ + 𝑖𝑛 − 1 Now, we can cancel ''1'' in right side of equation, we get (1 + 𝑖)𝑛 − 1 = 𝑛𝑖 + 𝑛(𝑛−1) 2 𝑖 2! +⋯ (1 + 𝑖)𝑛 − 1 = 𝑖 (𝑛𝑖 + (1+𝑖)𝑛 −1 ( = 𝑖 (1+𝑖)𝑛 −1 𝑖 𝑛(𝑛−1) 2 𝑖 2! 𝑛(𝑛−1) 2 𝑖(𝑛𝑖+ 𝑖 +⋯ ) 2! 𝑖 ) = (𝑛 + 𝑛(𝑛−1) 2! + ⋯ ) then = (𝑛 + 𝑛(𝑛−1) 𝑖+ 2! 𝑛(𝑛−1)(𝑛−2) 2 𝑖 3! 𝑖+ 𝑛(𝑛−1)(𝑛−2) 2 𝑖 3! + ⋯) +⋯) we can take common ''n'' in right side of equation, we get ( ( ( (1+𝑖)𝑛 −1 𝑖 (1+𝑖)𝑛 −1 𝑖 ) = 𝑛 (1 + (𝑛−1) 2! 𝑖+ (𝑛−1)(𝑛−2) 2 𝑖 3! 1 ) = 𝑛 (1 + ((𝑛 − 1)𝑖 + 2 (1+𝑖)𝑛 −1 𝑖 +⋯) (𝑛−1)(𝑛−2) 2 𝑖 2! 1 ) = 𝑛 (1 + (1 + (𝑛 − 1)𝑖 + 2 + ⋯ + 1 − 1)) (𝑛−1)(𝑛−2) 2 𝑖 2! 1 + ⋯)− ) 2 From step (1), we know (𝑛 − 1)(𝑛 − 2) 2 (1 + (𝑛 − 1)𝑖 + 𝑖 + ⋯ ) = (1 + 𝑖) 2! 𝑛−1 ( (1+𝑖)𝑛 −1 𝑖 1 (1+𝑖)𝑛−1 2 2 ) = 𝑛( + ) then 𝐹 = 𝐶( ⇒ 𝐹 𝑖 − = 2 ⇒ 2( 𝐹 𝑛𝐶 2𝐹 𝑛𝐶 1 (1+𝑖)𝑛−1 2 2 ) = 𝑛𝐶 ( + ) (1+𝑖)𝑛−1 1 𝑛𝐶 ⇒( (1+𝑖)𝑛 −1 2 1 − ) = (1 + 𝑖)𝑛−1 2 − 1) 1 (𝑛−1) =1+𝑖 1 (𝑛−1) 2𝐹 ⇒ 𝑖 = ( − 1) −1 𝑛𝐶 Answer: 𝑖=( 2𝐹 𝑛𝐶 − 1) 1 (𝑛−1) −1 Question #15 1. Given a complex valued function can be written as 𝑓(𝑧) = 𝑤 = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦) , where w is the real part of w and v is the imaginary part of v. Using algebraic manipulation figure out what wu and v is if 𝑓(𝑧) = 𝑧̅(1 − 𝑒 𝑖(𝑧+𝑧̅) ) Here 𝑧̅ denotes the complex conjugate of z. 2. Using the concept of limits figure out what the second derivative of f(z)=z(1-z) is. 3. Use the theorems of Limits that have been discussed before to show that lim [𝑥 + 𝑖(2𝑥 + 𝑦)] = 1 + 𝑖. [Hint: Use 𝑧 = 𝑥 + 𝑖𝑦] 𝑧⇒1−𝑖 Answer: Step 1 The given function is 𝑓(𝑧) = 𝑧̅(1 − 𝑒 𝑖(𝑧+𝑧̅) ) Substitute 𝑧 = 𝑥 + 𝑖𝑦 and 𝑧̅ = 𝑥 − 𝑖𝑦 in the above function and simplify as follows. Step 2 𝑓(𝑧) = (𝑥 − 𝑖𝑦)(1 − 𝑒 𝑖(𝑥+𝑖𝑦+𝑥−𝑖𝑦) ) = (𝑥 − 𝑖𝑦)(1 − 𝑒 𝑖2𝑥 ) = 𝑥 − 𝑖𝑦 − (𝑥 − 𝑖𝑦)𝑒 𝑖2𝑥 = 𝑥 − 𝑖𝑦 − (𝑥 − 𝑖𝑦)(cos(2𝑥) + 𝑖 sin(2𝑥)) = 𝑥 − 𝑖𝑦 − [𝑥 cos(2𝑥) + 𝑖𝑥 sin(2𝑥) − 𝑖𝑦 cos(2𝑥) + 𝑦 sin(2𝑥)] = 𝑥 − 𝑖𝑦 − 𝑥 cos(2𝑥) − 𝑖𝑥 sin(2𝑥) + 𝑖𝑦 cos(2𝑥) − 𝑦 sin(2𝑥) = [𝑥 − 𝑥 cos(2𝑥) − 𝑦 sin(2𝑥)] + 𝑖[−𝑦 − 𝑥 sin(2𝑥) + 𝑦 cos(2𝑥)] = 𝑢(𝑥, 𝑦) + 𝑖𝑣(𝑥, 𝑦) Thus, 𝑢(𝑥, 𝑦) = 𝑥 − 𝑥 cos(2𝑥) − 𝑦 sin(2𝑥) and 𝑣(𝑥, 𝑦) = 𝑦 cos(2𝑥) − 𝑦 − 𝑥 sin(2𝑥) Question #16 Let 𝑓(𝑡) = 𝑡 + 𝑖 where −𝜋 < 𝑡 < 𝜋 and has period 2𝜋. Why is it impossible to express the Fourier series of f(t) in real form? A. Because f(t) is a complex function with Re(f(x)) < 0 B. Because f(t) is a complex function with Im(f(x)) ≠ 0 C. Because f(t) is a complex function with Im(f(x))=0 D. Because f(t) is a complex function with Re(f(x)) > 0 Answer: Step 1 Given function is f(t)=t+i , −𝜋 < 𝑡 < 𝜋 Step 2 We see that 𝐼𝑚(𝑓(𝑡)) = 1 ⋯ (𝐼𝑚(𝑓(𝑡)) = coefficient of 𝑖) ⇒ 𝐼𝑚(𝑓(𝑡)) ≠ 0 Thus , f(t) is complex function with 𝐼𝑚(𝑓(𝑡)) ≠ 0 ⇒ 𝐼𝑚(𝑓(𝑥)) ≠ 0 ⇒ The Fourier series of f(t) cannot be expressed in real form. \therefore option (B) is correct Question #17 Find the complex zeros of the following polynomial function. Write f in factored form. 𝑓(𝑥) = 𝑥 3 − 12𝑥 2 + 49𝑥 − 58 The complex zeros of f are =? Answer: Step 1 Given: 𝑓(𝑥) = 𝑥 3 − 12𝑥 2 + 49𝑥 − 58 It can be written as, 𝑓(𝑥) = 𝑥 3 − 2𝑥 2 − 10𝑥 2 + 20𝑥 + 29𝑥 − 58 𝑓(𝑥) = 𝑥 2 (𝑥 − 2) − 10𝑥(𝑥 − 2) + 29(𝑥 − 2) 𝑓(𝑥) = (𝑥 − 2)(𝑥 2 − 10𝑥 + 29) Step 2 Roots of f(x) or zeroes of f(x) are given by (𝑥 − 2)(𝑥 2 − 10𝑥 + 29) ⇒ 𝑥 = 2 of 𝑥 2 − 10𝑥 + 29 = 0 𝑥 = 2 of 𝑥 = 10±√100−4(29) 2 = 5 ± 2𝑖 (complex zeroes) Question #18 𝑋2 Suppose X is a random variable such that E(3X-7)=8 and 𝐸 ( ) = 19. 2 What is Var(70-2X)? Answer: We have, E(3X-7)=8 3E(X)-7=8 3E(X)=15 15 𝐸(𝑋) = = 5 3 Also, 𝑋2 𝐸 ( ) = 19 2 𝐸(𝑋 2 ) = 19 × 2 = 38 Then, 𝑉(𝑋) = 𝐸(𝑋 2 ) − 𝐸 2 (𝑋) = 38 − 52 = 38 − 25 = 13 𝑉(70 − 2𝑋) = 𝐸(70 − 2𝑋)2 − 𝐸 2 (70 − 2𝑋) 𝐸(70 − 2𝑋)2 = 𝐸[702 − 2 × 70 × 2𝑋 + 4𝑋 2 ] = 702 − 280𝐸(𝑋) + 4𝐸(𝑋 2 ) 𝐸(70 − 2𝑋) = 70 − 2𝐸(𝑋) Then, 𝐸 2 (70 − 2𝑋) = 702 + 4𝐸 2 (𝑋) − 2 × 70 × 2𝐸(𝑋) = 702 − 280𝐸(𝑋) + 4𝐸 2 (𝑋) ⇒ 𝑉(70 − 2𝑋) = 702 − 280𝐸(𝑋) + 4𝐸(𝑋 2 ) − (702 − 280𝐸(𝑋) + 4𝐸 2 (𝑋)) = 4[𝐸(𝑋 2 ) − 𝐸 2 (𝑋)] = 4 × 𝑉(𝑋) = 4 × 13 = 52 Question #19 Let 𝑋1 , 𝑋2 , ⋯ , 𝑋𝑛 be n independent random variables each with mean 100 and standard deviation 30. Let X be the sum of these random variables. Find n such that 𝑃𝑟(𝑋 > 2000) ≥ 0.95 Answer: 𝑋1 , 𝑋2 , ⋯ , 𝑋𝑛 are independent ranodom variables with mean 100 and standard deviation 30 So, 𝑉𝑎𝑟(𝑋𝑖 ) = 302 = 900 Now given that, 𝑋 = ∑𝑖 𝑋𝑖 So 𝐸(𝑋) = 𝐸(∑𝑖 𝑋𝑖 ) = ∑𝑖 𝐸(𝑋𝑖 ) = 100𝑛 and 𝑉𝑎𝑟(𝑋) = 𝑉𝑎𝑟(∑𝑖 𝑋𝑖 ) = ∑𝑖 𝑉𝑎𝑟(𝑋𝑖 ) [ Covariances are zero due to independence] So, 𝑉𝑎𝑟(𝑋) = 900𝑛 By Central Limit Theorem, 𝑋−𝜇 𝑋−100𝑛 ≡ ∼ 𝑁(0,1) 𝜎 √900𝑛 Now, P(X>2000)>0,95 implies 1 − 𝑃(𝑋 ≤ 2000) > 0,95 𝑃(𝑋 ≤ 2000) < 0,05 𝑋 − 100𝑛 2000 − 100𝑛 𝑃( ≤ ) < 0,05 30√𝑛 √900𝑛 2000−100𝑛 implies Φ [ ] < 0,05 = Φ(−1,645) implies 30√𝑛 2000−100𝑛 30√𝑛 < −1,645 Here you can solve this in two methods, Trial and error method Numerically using root finding method Using trial and error method, For n = 22 2000−100𝑛 = −1,4213 30√𝑛 For n=23 2000 − 100𝑛 = −2,085 30√𝑛 So, for n = 23 it is crossing -1.645. Hence n = 23 Question #20 Suppose that the random variables X and Y have the joint p.d.f. kx( x − y),0  x  2 , − x  y  x f ( x, y) =  ,elsewhere 0 (i) Evaluate the constant k. (ii) Find the marinal p.d.f. of the two random variables Answer: Step 1 Consider the given function. kx( x − y),0  x  2 , − x  y  x f ( x, y) =  ,elsewhere 0 Step 2 For part (i) it is required to determine the value of constant k. f(x,y) is a valid pdf then, ∫𝑥 ∫𝑦 𝑓(𝑥, 𝑦)𝑑𝑦𝑑𝑥 = 1 So, 2 𝑥 ∫0 ∫−𝑥 𝑘𝑥(𝑥 − 𝑦)𝑑𝑦𝑑𝑥 = 1 2 𝑥 𝑘 ∫0 ∫−𝑥(𝑥 2 − 𝑥𝑦)𝑑𝑦𝑑𝑥 = 1 2 𝑘 ∫0 [𝑥 2 𝑦 − 𝑥𝑦 2 𝑥 ] 𝑑𝑥 = 1 2 −𝑥 2 𝑥3 3 𝑘 ∫0 [𝑥 − − (−𝑥 3 2 2 3 ]𝑑𝑥 [2𝑥 𝑘 ∫0 =1 2 𝑥4 − 𝑥3 2 )] 𝑑𝑥 = 1 𝑘[ ] = 1 2 0 𝑘 2 [16] = 1 𝑘= 1 8 Step 3 For part (ii) it is required to determine marginal pdf of the two random variables.  x( x − y ) 0  x  2, − x  y  x  f ( x, y ) =  8  , elsewhere 0 𝑥 𝑥 𝑥(𝑥−𝑦) 𝑓𝑋 (𝑥) = ∫−𝑥 𝑓(𝑥, 𝑦)𝑑𝑦 = ∫−𝑥 1 𝑓𝑋 (𝑥) = [𝑥 2 𝑦 − 8 8 𝑥 𝑥𝑦 2 2 𝑑𝑦 ] 1 −𝑥 = [2𝑥 3 ] 8 𝑓𝑋 (𝑥) = 𝑓𝑌 (𝑦) = 1 𝑥3 = [ 8 3 1 8 𝑥3 4 2 ∫0 𝑓(𝑥, 𝑦)𝑑𝑦 2 𝑥2𝑦 − 2 ] 0 = [ − 2𝑦] 8 3 1 𝑦 3 4 𝑓𝑌 (𝑦) = − 2 𝑥(𝑥−𝑦) = ∫0 8 𝑑𝑥