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Solutions to Homework Set 1

1. As |i| = 1 and Arg(i) = /2, the absolute value of all fourth roots ofimust be 1, and their argument has to be such that 4 = (/2) +2kfor some integer p. Solving this last equation, for , we get that =

(/8 ) + (k/2) for some integer k. This gives us for different roots, fork = 0, 1, 2, 3, after which we will not get new roots since the new valuesof will be 2k larger than the old values of.

This means that the equation x4 = i has four solutions, namely x1 =ei/8, x2 = e

5i/8, x3 = e9i/8 x2 = e

13i/8.

2. We have

x8 + 4 = x8 (4) = (x4 + 2i)(x4 2i) = 0.

So the roots are the four roots ofx4 = 2i and the four roots ofx4 = 2i.

These can be found just as in the previous problem, except that theyall have absolute value 21/4 instead of 1.

In particular, the four roots ofx4 = 2i are the four roots of the previousproblem multiplied by 21/4, while the four roots ofx4 = 2i are equalto the four roots ofx4 = 2i times ei/4, since the argument of2i is more than that of 2i.

There are several ways to write x8 +4 as a product of polynomials withreal coefficients. We will discuss this in detail in class. For now, justnote that

x8

+4 = (x4

+2)2

4x4

= (x4

+2)2

(2x2

)2

= (x4

2x2

+2)(x4

+2x2

+2).

3. Let the product we are looking for be denoted by Pn. Then we claimthat Pn = 1 ifn is odd, and Pn = 1 if n is even. Indeed, if u is aroot of unity, then so is 1/u. So ifu shows up in our product, so does1/u, so they cancel each other. So every term of our product cancelexcept the term that is equal to its own reciprocal, since in that case,u = 1/u does not show up one more time. The equation u = 1/u hastwo solutions, u = 1, and u = 1. However, ifn is odd, then 1 is nota root of unity, so that does not show up in our product. So for oddn, the only term that does not cancel is 1, while for even n, the onlyterms that do not cancel are 1 and 1.

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4. Let u be the 15th root of unity that has the smallest positive argument.Then all other roots of unity are of the form ui, where 1 i 15.Clearly, ui is a primitive 15th root of unity if and only if i is notdivisible by 3 or 5.

So we are looking for

u+u2+u4+u7+u8+u11+u13+u14 =15

i=1

ui5

j=1

(u3)j3

j=1

(u5)j +1 = 1.

Indeed, out of the three summation signs, the first one sums all fifteenthroots of unity, the second one sums all fifth roots of unity, and the thirdsums all fifth roots of unity. Each of these sums is 0 as we learned. Thedifference between these three terms and the left-hand side is 1, sincethe first sum contains 1 with a positive sign, but the second and thirdterms contain 1 with a negative sign. This completes the proof.

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