AMS10 HW1 Grading Rubric

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1 AMS10 HW1 Grading Rubric Problem 1 (16ts- ts/each). Left hand side is shown to equal right hand side using examles with real vectors. A vector sace is a set V on which two oerations, vector addition and scalar multilication are defined. Construct an examle with real vectors for each of the following conditions and show them to be true. The vector addition oeration (+) must satisfy the following conditions: (1) Commutative law: For all vectors ~u and ~v in V, ~u + ~v = ~v + ~u () Associative law: For all vectors ~u,~v, and ~w in V, ~u +(~v + ~w )=(~u + ~v )+ ~w () Additive identity: The set V contains an additive identity element, denoted by a 0, such that for any vector ~v in V, ~u + 0 = 0 + ~u (4) Additive inverses: For each vector ~v in V, the equations ~v + ~x = 0 and ~x + ~v = 0 have a solution ~x in V, called an additive inverse of ~v, and denoted by ~v. Note: Closure: If~u and ~v are any vectors in V, then the sum ~u + ~v belongs to V. The scalar multilication oeration ( ) is defined between real numbers (or scalars) and vectors, and must satisfy the following conditions: (5) Distributive law: For all real numbers c and all vectors ~u, ~v in V, c (~u + ~v )=c ~u + c ~v (6) Distributive law: For all real numbers c, d and all vectors ~v in V, (c + d) ~v = c ~v + d ~v (7) Associative law: For all real numbers c, d and all vectors ~v in V, c (d ~v )=(cd) ~v (8) Unitary law: For all vectors ~v in V, 1 ~v = ~v Note: V. Closure: If~v is any vector in V, and c is any real number, then the roduct c ~v belongs to Problem (5ts). Draw the vector ~a ~ b using the arallelogram law. Note that ~a ~ b = ~a +( ~ b), 4

2 ~a ~ b ~a ~ b ~ b 5

3 Problem (18 ts). a.(4ts - ts each) Calculate the absolute value z for the following comlex numbers i. z = 1 + i ii. z =5 i. z =1 + i5 ii. z =5 b. (4ts - ts each) Write down the comlex exonential form of i. z = 1 + i ii. z =5 + i5 (Hint: there are an infinite number of reresentations) i. z = e i( + k) for k =0, ±1, ±,... ii. z =5e i( 4 + k) for k =0, ±1, ±,... c. (10 ts)use the comlex exonential form of z = 10 has only two distinct solutions. We have +i10 1 to show that x = 10 +i10 1 z = 10 + i101 (11) z = 10e i( 6 + k) (10ts if you got this far) (1) (z ) 1/ = (10e i( 6 + k) ) 1/ (1) z = 10e i( 1 + k) (14) for k =0, ±1, ±,... Note that 10e i( 1 + ) = 10e i( 1 + ) = 10e i( 1 +5 ) = 10e i( 1 +(k+1) ) and 10e i( 1 + ) = 10e i( 1 +4 ) = 10e i( 1 +6 ) = 10e i( 1 +k ) and so there are only two unique solutions (full credit). Problem 4 (15ts-ts/each). Simlifying fractions of comlex numbers. Suose we have a comlex number exressed as the division of two distinct comlex numbers of the form z = a 1 + ib 1 a + ib. 6

4 Then a method of finding the real and imaginary comonent involves the comlex conjugate of the denominator in order to remove any imaginary numbers from the denominator. Recall that z z = z is a real non-negative number. The following is an illustration of this z = a 1 + ib 1 a + ib (15) = 1 1 a1 + ib 1 a + ib (16) = a ib a ib a1 + ib 1 a + ib (17) = (a ib )(a 1 + ib 1 ) a +. (18) b Aly this method to find the real and imaginary comonents of the following exressions (don t just lug in values but show work): +i a. +i b. 1+i i c. 1 i +i 1 i d. 1+i (19) (0) (1) () () a. +i +i = i i +i 6+i i +1 = = 7 i = 7 1 +i i (4) b. 1+i i = +i +i 1+i i = +i 1 = 1+i = i (5) c. 1 i +i = i i 1 i 4i i 4 = = 6i = 1 +i i (6) d. 1 i 1+i = 1 i 1 i 1 i 1+i = 1+i i = 5 5 = 1 (7) (8) Problem 5 (18 ts- ts/each). Division of comlex numbers using the exonential form. Suose you are given two comlex numbers in the form z 1 = r 1 e i 1 and z = r e i, then the division is given by z 1 = r 1 e i 1 z r e i = r 1 e i( 1 ) r which is equivalent to z 1 z = r 1 r (cos( 1 )+i sin( 1 )). i. Find the argument (or hase) for the numerator and denominator of (c) and (d) in roblem 4. Find the angle using arctan and alying the aroriate adjustment. Show work. Verify the answer using the Matlab command angle (e.g. angle(1+i)). 7

5 c. arg(1 i) = arctan( ) (9) arg( + i) = arctan(1) (0) d. arg( 1 i) = arctan() (1) arg(1 + i) = arctan() () ii. Find the modulus (or absolute value) for the numerator and denominator of (c) and (d) in Problem 4. You may work out by hand or use the Matlab command abs (e.g. abs(1+i)). c. 1 i = 5 () +i = 8 (4) d. 1 i = 5 (5) 1+i = 5 (6) iii. Using the results from (a) and (b), write the comlex exonential form for the numerator and denominator of (c) and (d) in Problem 4. Calculate the real and imaginary comonent of (c) and (d) by dividing the comlex exonential functions. Verify that the final comlex exonential form (z 1 /z = re i = cos( )+isin( )) is equivalent to the results found in Problem 4. Recall, two comlex numbers a + ib and c + id are equal if and only if a = c and b = d. Alying the definition of the comlex exonential we find that the real art is c. Re: r 1 r cos( 1 )= Im: r 1 r cos( 1 )= 5 8 cos(arctan( ) arctan(1)) =.5 (7) 5 8 sin(arctan( ) arctan(1)) =.75 (8) d. Re: r 1 r cos( 1 ) = cos( ) = 1 (9) Im: r 1 r cos( 1 )=sin( ) =0 (40) Problem 6 (9ts-ts/each). Given the comlex number z = roots (z 1/m )for a. m=10 b. m=45 c. m=100 It is ok to leave in exonential form. The comlex form is z = 5e i where = arctan( ) +. 1+i find the m distinct m-th 8

6 a. 5 1/0 e i( + k)/10 for k =0, 1,...,9 (41) b. 5 1/90 e i( + k)/45 for k =0, 1,...,44 (4) c. 5 1/00 e i( + k)/100 for k =0, 1,...,99 (4) (44) Problem 7 (9ts-ts/each). Calculate the following owers of z = 1+i a. z 10 b. z 15 c. z 00 It is ok to leave in exonential form. Solution a. 5 5 e i 10 (45) b. 5 15/ e i 15 (46) c e i 00 (47) (48) Problem 8. Construct the following vectors in Matlab: a. A x1 vector with all 1 s in the entries and define it u. b. A x1 vector with the first element 4 and the second element and define it v c. What is 5u+v? Verify using Matlab Write down using Matlab syntax the exact exression you would tye into the command window for a-c. a. u=[1;1]; b. v=[4;]; c. 5*u+v, answer: 5u+v=[9;8] *1ts for submission 9