lina-16-10-09-p1 - z = x i y we defined ¯ z:= x-i y and |...

Info icon This preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Linear Algebra & Geometry: Sheet 2 Set on Friday, October 16: Questions 1, 2, 4 (a)(c)(e), 5 (i), 6 1. Consider the following vectors: ( a ) 1 1 ( b ) 0 - 5 ( c ) - 3 4 , find for each of them a λ > 0 and a θ [0 , 2 π ) such that v = λ u ( θ ) where u ( θ ) = cos θ sin θ . 2. Find the components of the vector x R 2 that satisfies x = λ u ( θ ) for (a) λ = 1 and θ = π/ 3 (b) λ = 2 and θ = π/ 2 (c) λ = 10 and θ = 7 π 6 3. Let v 1 , v 2 , · · · , v n R 2 be n arbitrary vectors in R 2 . Show that k v 1 + v 2 + · · · + v n k ≤ k v 1 k + k v 2 k + · · · + k v n k and give an example of n vectors for which there is equality. 4. Compute the following expressions, i.e., write them in the form a + i b with explicit numbers a and b , ( a ) (1 + i) + (2 - 3i) ( b ) (1 + i)(1 + 2i) - 3 ( c ) 1 1 - i ( d ) 2 + i 1 + i ( e ) (2 - 3i)(3 - 2i) 1 + i ( f ) 1 (1 + i)(2 - i) 5. Recall that for a complex number
Image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: z = x + i y we defined ¯ z := x-i y and | z | = √ ¯ zz . Show that (i) ( a ) z 1 + z 2 = ¯ z 1 + ¯ z 2 ( b ) z 1 z 2 = ¯ z 1 ¯ z 2 ( c ) z 1 /z 2 = ¯ z 1 / ¯ z 2 ( d ) ¯ ¯ z = z (ii) ( e ) | z 1 z 2 | = | z 1 || z 2 | ( f ) | z 1 + z 2 | ≤ | z 1 | + | z 2 | 6. (i) Find x,y ∈ R such that ( a ) e i π 3 = x + i y ( b ) 2e i π 2 = x + i y ( c ) 10e i 7 π 6 = x + i y and compare with Question 2 . (ii) Find r ∈ R + , ϕ ∈ [0 , 2 π ) such that ( a ) 1 + i = r e i ϕ ( b )-5i = r e i ϕ ( b )-3 + 4i = r e i ϕ and compare with Question 1 . 1...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern