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QUIZ/MIDTERM #1 SOLUTIONS (1) For each of the following equations, find all complex z which satisfy the equation. Your answer must be in either polar or rectangular form. (a) z 3 + 7i = 0. (b) |ez | = e2 . (This is different than the problem you saw on homework!) (c) 1 + z 2 + z 4 + z 6 = 0. Solution. (a) We want to solve z 3 = −7i. Use polar coordinates: −7i = 7e3πi/2 , so if z = reiθ , then r3 = 7, 3θ = 3π/2 + 2nπ, where n is any integer.√Therefore θ = π/2, π/2 + 2π/3 = 7π/6, π/2 +√4π/3 = 11π/6, r = 3 7, which √ √ 3 3 πi/2 give solutions z = 7e , 7e7πi/6 , 3 7e11π/6 . A fairly common minor √ error was to let z = − 3 7eπi/6 , say. Strictly speaking, in polar form, the negative sign should be absorbed into the eiθ term, since r > 0. (b) If z = x + iy, then ez = ex eiy expresses ez in polar form. Therefore |ez | = ex (alternately, |ez | = |ex ||eiy | = |ex | = ex , because |eiy | = 1 for real y, and ex > 0 is always true). Therefore |ez | = e2 if and only if z = 2 + iy, where y is any real number. (c) There are a variety of ways to solve this problem. One way is to notice that the polynomial 1 + z 2 + z 4 + z 6 = (1 + z 4 )(1 + z 2 ) = 0. The solutions to 1+z 2 = 0 are z = ±i. To solve 1+z 4 = 1, use polar coordinates: if z = reiθ , then (reiθ )4 = r4 e4iθ = 1 · eπi , so r = 1, θ = π/4, 3π/4, 5π/4, 7π/4. Therefore z = eπi/4 , e3πi/4 , e5πi/4 , e7πi/4 solve z 4 = 1, and together with z = ±i give all solutions to this equation. Alternately, one might recall that we can factor 1 + z 2 + z 4 + z 6 = (1 − z 8 )/(1 − z 2 ). This is equal to 0 if z 8 = 1, z 2 6= 1, which means z can be any 8th root of unity except for ±1. (2) Let x, y be real numbers. Show that cos(x + y) = cos x cos y − sin x sin y and sin(x + y) = sin x cos y + sin y cos x. (Hint: Think polar!) Solution. If x, y are real numbers, then eix = cos x + i sin x, eiy = cos y + i sin y. Then eix · eiy = ei(x+y) = cos(x + y) + i sin(x + y). On the other hand, we can also express this product as eix · eiy = (cos x + i sin x)(cos y + i sin y) = (cos x cos y − sin x sin y) + (sin x cos y + cos x sin y)i. Equating real and imaginary parts of these two expressions proves the two identities in the problem. (3) Let a, b, c, d ∈ R be real numbers such that ad − bc = 1. Let H be the upper half-plane {z ∈ C| Im z > 0}. Show that if z ∈ H, then az + b ∈H cz + d 1 2 QUIZ/MIDTERM #1 SOLUTIONS as well. (Hint: Don’t think polar!) This calculation appears in the study of modular forms, which is a large branch of modern number theory. Solution. We want to show that the imaginary part of (az + b)/(cz + d) is greater than 0. To do this, we eliminate complex numbers from the denominator by multiplying the numerator and denominator by the complex conjugate of the denominator: az + b az + b cz + d (az + b)(cz + d) = · = , cz + d cz + d cz + d |cz + d|2 where we used various simple properties of complex conjugation in the equation above, together with the fact that c, d are real numbers. Notice that |cz + d|2 > 0: indeed, we know that |cz + d|2 ≥ 0 is always true, and also that cz + d 6= 0, since if cz + d = 0, then either z = −d/c or c, d = 0. The former is impossible since z ∈ H, which means z has positive imaginary part, and the latter is impossible since ad − bc = 1. (Virtually everyone missed checking that cz + d 6= 0.) Therefore to solve the problem it suffices to show that (az + b)(cz + d) has positive imaginary part. We have (az + b)(cz + d) = ac|z|2 + bd + adz + bcz. Notice ac|z|2 , bd are both real, so Im (az + b)(cz + d) = Im (adz + bcz). If we let z = x + yi, then Im (adz + bcz) = ady − bcy = (ad − bc)y = y > 0, as desired. (4) It is easy to show that a finite union of closed sets in C will also be closed in C. In this problem you will show that, in general, one cannot say anything about an infinite union of closed sets. (a) (15) Give an example of an infinite collection of closed sets in C whose union is closed. (b) (15) Give an example of an infinite collection of closed sets in C whose union is not closed. Solution. Both parts of this problem have many possible different answers. We provide a few simple possibilities here. In both parts we use the basic fact that a set consisting of a single point in C is closed. (a) The most trivial possible example is to take an infinite collection of closed sets all equal to the same closed set X. Then their union is X, which is closed. For a slightly less trivial answer, let X be any closed set, and then take the collection consisting of single point sets {x}, where x ranges over points in X. Then the union of these sets is X, which is closed. (b) In line with the previous solution, perhaps the simplest solution is to let X be any non-closed set (say, an open disc, although it doesn’t matter), and take the collection to be the single point sets {x}, x ∈ X. Then the union of this collection is X, which is not closed. A fairly common error was to choose, say, closed discs of radius n, as n → ∞. These sets have union all of C, but C is a closed (as well as open) set, so is not a correct answer to this problem. Of course, this example would have worked on the previous part.