1873_solutions

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

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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  n1  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...
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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.

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