1873_solutions

All x then we ought to conclude that ought to be it

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

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ots of 8? 3 i. Since the equation The equation z 2  2z  4  0 says that z  1 2  3 which yields z  1 also implies that z 2 z 2  2z  4  0 which says that z 3 8  0, the solutions of the equation must be cube roots of 8. 3. Given two complex numbers z and w with w similar identity for absolute value. 0, prove that the complex conjugate of z/w is z/w. Establish a Hint: Once the corresponding assertion for multiplication has been obtained we can obtain the assertion for division very quickly using the fact that z z  w w. 4. Work out 1  i n for a few positive integers n. What is 1  i 4 ? 5. Prove that if n is an even natural number that is not a multiple of 4 then the real part of 1  i n is zero. Hint: Use mathematical induction. We want to prove that whenever n is a positive integer, the real part of the number 1  i 4n 2 is zero. This fact is obvious if n  1. Now given any positive integer n for which the real part of the number 1  i 4n 2 is zero we have 1  i 4 n 1 2  1  i 4 n 2 1  i 4  4 1  i 4 n 2 and therefore this number has a zero real part. It follows from the principle of mathematical induction that 1  i 4n 2 has a zero real part for every positive integer n. 6. Prove that if n is a natural number then 1  i n is real and positive if and only if n is a multiple of 8. If n is any positive integer then, since 1  i 4n  1  i 4n 2 1  i 2  2i 1  i 4n 2 and since the number 1  i 4n 2 is imaginary, we deduce that 1  i 4n is real. We deduce that if n is any even positive integer then the number 1  i n must either be real or imaginary and it follows that the number 1  i n 1  1  i n  i 1  i n cannot be real. We conclude that if n is an odd positive integer then 1  i n cannot be real. Finally, if n is a positive integer then since n 1  i 4n  1  i 4  4n we see that the number 1  i 4n is positive if and only if n is even. Therefore the number 1  i real and positive if and only if n is a multiple of 8. n is 7. Assuming the standard identities from trigonometry, prove that if  and  are any real numbers then we have cos   i sin  cos   i sin   cos     i sin    . Then prove by induction (or otherwise) that if n is any natural number then cos   i sin  n  cos n  i sin n. 93 Does the same identity hold if n is a negative integer? The first equation follows simply: cos   i sin  cos   i sin   cos  cos  sin  sin   i sin  cos   sin  cos   cos     i sin    . The equation cos   i sin  n  cos n  i sin n is obvious when n  1. Now given any positive integer n for which the equation cos   i sin  n  cos n  i sin n holds we have cos   i sin  n1  cos   i sin  n cos   i sin   cos n  i sin n cos   i sin   cos n    i sin n   and so it follows from the principle of mathematical induction that the equation cos   i sin  n  cos n  i sin n holds for every positive integer n. Since cos  i sin  1  cos   i sin  1  cos   i sin  cos   i sin...
View Full Document

This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.

Ask a homework question - tutors are online