{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

math 375 solution - Solutions to Exam 1 Review Problems 2 1...

Info icon This preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Solutions to Exam 1 Review Problems 1. Let z = - 2 1 - 3 i . (a) Calculate the rectangular forms of z , z , and z - 1 . (b) Calculate | z | , Arg( z ) and the polar form of z . (c) Calculate the rectangular form of z 42 . ( Hint: Use your answer from part (b).) Solution. (a) z = - 1 2 - i 3 2 z = - 1 2 + i 3 2 z - 1 = z = - 1 2 + i 3 2 ( z - 1 6 = z always, it’s only true whenever | z | = 1) (b) Calculate | z | , Arg( z ) and the polar form of z . | z | = 1 Arg( z ) = - 2 π 3 z = e - i 2 π 3 (c) z 42 = e - i 2 π 3 42 = e - i 2 π 3 · 42 = e - i 28 π = e i ( - 14)2 π = 1 2. Solve the equation | e - 1 | = 2 for θ real. Solution. | e - 1 | = 2 | e - 1 | 2 = 4 ( e - 1 ) ( e - 1) = 4 ( e - 1 ) e - 1 = 4 ( e - 1 ) ( e - - 1 ) = 4 1 - e - e - + 1 = 4 e + e - = - 2 e + e - 2 = - 1
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
cos( θ ) = - 1 θ = (2 k + 1) π 3. Prove that 1 e = e - , where φ is a real number. Solution. 1 e = 1 cos( φ ) + i sin( φ ) = 1 cos( φ ) + i sin( φ ) · cos( φ ) - i sin( φ ) cos( φ ) - i sin( φ ) = cos( φ ) - i sin( φ ) cos 2 ( φ ) + sin 2 ( φ ) = cos( φ ) - i sin( φ ) = cos( - φ ) + i sin( - φ ) = e i ( - φ ) 4. Find all the cube roots of w = 3 3 + 3 i . Solution. In polar form, 3 3 + 3 i = 6 e iπ/ 6 , so, writing z = re we want z 3 = r 3 e i 3 θ = 6 e iπ/ 6 For this we need that r 3 = 6 and 3 θ - π 6 = 2 or r = 3 6 and θ = (12 k + 1) π 18 So the cube roots of 3 3 + 3 i all look like 3 3 e i (12 k +1) π 18 . However, among the infinitely many values for k there are only three distinct roots n e iπ/ 18 , e i 13 π/ 18 , e i 25 π/ 18 o 5. Prove that 1 z 2 - 1 1 3 for all z on the circle of radius 2 centered at the origin. What can you say about 1 z 2 - 1 on the circle of radius R centered at the origin? Solution. By the reverse triangle inequality | z 2 - 1 | ≥ | z 2 | - | 1 | = | z | 2 - 1 for any z . Thus, for z on the circle | z | = 2 we have, | z 2 - 1 | ≥ 2 2 - 1 = 3, so 1 z 2 - 1 1 3 Similarly, for any R > 1 and all z on the circle | z | = R , 1 z 2 - 1 1 R 2 - 1
Image of page 2
6. Solve each of the following equations or explain why no solution exists. (a) z 3 = - 27 (b) e z = 2 - 2 i (c) Log z = 2 πe iπ/ 6 (d) z 1 / 3 = i Solution. (a) This is solved exactly like number 4 using the polar form - 27 = 37 e . The solutions are 3 e iπ/ 3 , 3 e , 3 e i 5 π/ 3 (b) In polar form 2 - 2 i = 8 e - iπ/ 4 so, writing e z = e x e iy , we want e x e iy = 8 e - iπ/ 4 For this we need that e x = 8 and y - - π 4 = 2 or x = ln 8 and y = - π 4 + 2 = (8 k - 1) π 4 So the set of solutions of e z = 2 - 2 i is ln 8 + i (8 k - 1) π 4 | k Z (c) Using the fact that exp(Log( z )) = z for all z we can solve the equation Log( z ) = 2 πe iπ/ 6 as follows z = exp(Log( z )) = exp 2 πe iπ/ 6 = exp (2 π (cos( π/ 6) + i sin( π/ 6)) = exp 2 π 3 2 + i 1 2 !! = exp 3 π + = e 3 π (cos( π ) + i sin( π )) = e 3 π ( - 1 + i · 0) = - e 3 π (d) This one is tricky. By definition, for any z , z 1 / 3 = exp 1 3 Log( z ) = | z | 1 / 3 e i Arg( z ) / 3 Now, - π < Arg( z ) π , so - π/ 3 < Arg( z ) / 3 π/ 3. So, for any z , z 1 / 3 must lie in the polar region { ( r, θ ) | r 0 , - π/ 3 < θ π/ 3 } . Since i is not in this region, the equation z 1 / 3 = i has no solution.
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
7. Write the function f ( z ) = z 3 + z + 1 in the form f ( z ) = u ( x, y ) + iv ( x, y ).
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the 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