ps03sol

ps03sol - 18.03 Problem Set 3 Fall 2009 Solutions 1(5...

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Unformatted text preview: 18.03 Problem Set 3 Fall 2009 Solutions 1. (5 points) Solve the equation x 6- 1 = 0 using just high school algebra (factorization and the quadratic formula). Then solve it using DeMoivre’s formula. Say how the six roots that you find by each method match up (for instance, by explaining why e 5 iπ/ 6 =- 1 , or whatever happens to be true). We have x 6- 1 = ( x 3- 1)( x 3 + 1), as it is the difference of two squares. Note that 1 is a root of x 3- 1 and- 1 is a root of x 3 + 1, and so we can factor out x- 1 and x + 1 from these terms: x 6- 1 = ( x 3- 1)( x 3 + 1) = ( x- 1)( x 2 + x + 1)( x + 1)( x 2- x + 1) . Using the quadratic formula, the roots of the two quadratic terms are 1 2 (- 1 ± i √ 3) and 1 2 (1 ± i √ 3), respectively. These four roots, along with ± 1, are the 6 solutions of x 6- 1 = 0. Suppose re iθ is a root. Then, by DeMoivre’s formula, 0 = ( re iθ ) 6- 1 = r 6 e i 6 θ- 1 , and so r 6 e i 6 θ = 1. The norm of 1 is 1, and so r = 1. Since e i = 1, we must have 6 θ = 0 + 2 kπ for some whole number k , i.e., θ is a multiple of π/ 3. The first 6 multiples lead to distinct powers, and line up with our previous solutions as follows: e iπ/ 3 = 1 2 + √ 3 2 i e i 2 π/ 3 =- 1 2 + √ 3 2 i e i 3 π/ 3 =- 1 e i 4 π/ 3 =- 1 2- √ 3 2 i e i 5 π/ 3 = 1 2- √ 3 2 i e i 6 π/ 3 = 1 . 2. (5 points) A “natural logarithm” of a complex number w is another complex number z so that w = e z . When this is true, we write z = log( w ); care is required since log( w ) is not just one complex number but many. a) Find a natural logarithm of 1- i √ 3 . We know that z = log(1- i √ 3) just when e z = 1- i √ 3. Suppose z = a + bi , with a and b real. Then 1- i √ 3 = e z = e a + bi = e a e bi , Where e a is real, and e bi has norm 1. So e a must equal the norm of 1- i √ 3, i.e., p (1 2 + √ 3 2 ) = 2....
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ps03sol - 18.03 Problem Set 3 Fall 2009 Solutions 1(5...

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