MTH 3102 Complex Variables Practice Exam 1 Feb. 10, 2017

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1 Name (Last name, First name): MTH 310 Comlex Variables Practice Exam 1 Feb. 10, 017 Exam Instructions: You have 1 hour & 10 minutes to comlete the exam. There are a total of 7 roblems. You must show your work. Partial credit may be given even for incomlete roblems as long as you show your work. 1. Find the exonential form of ( + i) 6. Solution 1 First, 1 + i 1 = e i π 4 so that + i = ( 1 + i 1 ) = e i π 4. From this we conclude that ( + i) 6 = ( ) 6 ( e i π 4 = 6 ) e i 6π 4 = 6 3 e i 3π = 9 e i 3π = i 9. 1

2 . Draw vectorially the numbers (a) ( + i) ; (d) 11+ii. 3i; (b) 1 + 3i; (c) + i and Solution (a) We have 1 + 3i = 1 + i 3 = ei 3 ; (b) 1 + 3i = e 3 6 (c) + i = ei 4 and ( + i) = 9 ei = i9 ; (d) 1 i e i4 1+i = = i =e 1+i 1+i e 4 i = i: i3 ;

3 3. Find ( 8 8 3i ) 1/4, exress the roots in rectangular coordinates, exhibit them as the vertices of a certain square, and oint out which is the rincile root. Solution 3 First, we exress 8 8 3i in its olar form as 8 8 ( 3i = ( ) ) 1 3 3i = 16 + i = 16e i π 3 = 16e i( π 3 π) = 16e i π 3. Therefore, ( 8 8 ) 1/4 3i ( 4 = 16e i π +kπ 3 4 = 4e i( π 6 + kπ ) = 4e i π 6 e i( kπ 4 ) = 4e i π 6, 4e i π 6 i, 4e i π 6, 4e i π 6 i ) for k = 0, 1,, 3. The rincile root is when k = 0 here, i.e., rincile root: 4e i π 6. We can exress them in rectangular coordinates as ( 4e i π π ) ( π )) 6 = 4 (cos sin 6 6 ( ) 3 = 4 i1 = 3 i 4e i π 6 i = ( ) 3 i i = + i 3, 4e i π 6 = 3 + i, 4e i π 6 i = ( + i ) 3 = i 3. Here is a icture of all this: 3

4 4. State the de nition of a boundary oint of a set and the de nition of an oen set of C. Prove that a set S is oen if and only if each oint of S is an interior oint. Solution 4 Let S C. A oint z C is called a boundary oint of the set S if for every " > 0, the oen ball B (z; ") = fw C : jw zj < "g has oints in S and oints which are not in S. A set O C is said to be an oen set of C if it does not contain any of its boundary oints. Proof. Let S C. We will now rove that S is oen if and only if each oint of S is an interior oint. Suose each oint of S is an interior oint. Let z C be a boundary oint of S then z cannot be an interior oint for otherwise there exists " > 0 such that the oen ball B (z; ") lies in S contradicting the fact z is a boundary oint of S. Thus, S does not contain any of its boundary oints and hence S is an oen set. Conversely, suose that S is an oen set. Then S does not contain any of its boundary oints. Now let w S. Then there must exists an " > 0 such that B (z; ") lies in S for otherwise every neighborhood of w could contain oints not in S and oints in S (for examle, w) imlying w is a boundary oint, a contradiction that S does not contain any of its boundary oints. Thus, there is a neighborhood of w that is comletely contained in S imlying that w is an interior oint of S. Therefore, as w was an arbitrary element of S, this imlies each oint of S is an interior oint. This comletes 4

5 the roof. 5

6 5. Show that the limit of z3 1 z +1 Solution 5 To rove that we need only rove that as z is. z 3 1 lim z z + 1 =, lim z 0 [ ( 1 ) 3 z 1 ] 1 ( 1 ) = 0. z + 1 But the latter is true since lim z 0 [ ( 1 ) 3 ] 1 z 1 ( 1 ) = lim z + 1 z 0 ( 1 ) z + 1 ( 1 z ) 3 1 z + z 3 = lim z 0 z 3 = = 0. 6

7 6. For the function f (z) = z, find all oints z where: (a) it is not differentiable; (b) it is differentiable; (c) is a singular oint. Solution 6 First, f has domain C and on this domain we have f (z) = z = x + y, z = x + iy, x, y R. This means that is comonents (real and imaginary arts) u, v are u (x, y) = x + y, v (x, y) = 0. These functions are have continuous first-order artial derivatives on the whole domain and therefore by these facts and by the theorem in Sec. 3 on suffi cient conditions for a comlex-valued function to be diff erentiable, we need only find those oints in C where the Cauchy-Riemann equations are satisfied or not satisfied in order to find the set of oints where f is differentiable and the set of oints where f is not diff erentiable, resectively. Thus, since u x (x, y) = x, u y (x, y) = y, v x (x, y) = 0, v y (x, y) = 0 the Cauchy-Riemann equations for f are x = u x (x, y) = v y (x, y) = 0, y = u y (x, y) = v x (x, y) = 0. These are only satisfied when (x, y) = (0, 0), that is, when z = 0. Therefore, (a) {0} are the set of oints where f is differentiable, and (b) {z C : z 0} are the set of oints where f is not differentiable. Moreover since, for any z C with z 0, every unctured neighborhood of z has only oints where f is not diff erentiable, then (c) f has no singularities. 7

8 7. (a) Prove that for the rincial branch of the logarithm, f (z) = Log z, it is differentiable in its domain and find its derivative; (b) Find Log (1 + i) and for the multivalued logarithm, log z, find log (1 + i) in terms of Log (1 + i); (c) Are the comonents of the f (z) harmonic in its domain? Proof. (a) The function which has domain f (z) = Log z = ln z + i Arg z, D = {z C : z 0 and π < Arg z < π} is single-valued. Furthermore, in its olar form it has comonents f ( re iθ) = ln r + iθ, (r > 0, π < θ < π) u (r, θ) = ln r, v (r, θ) = θ which have continuous first-order artial derivatives on its domain so that by the theorem in Sec. 4, it will be differentiable on its domain if the olar form of the Cauchy-Riemann equations are satisfied everywhere, which they are since and its derivative is then given by i.e., (b) We have And since then ru r (r, θ) = 1 = v θ (r, θ), u θ (r, θ) = 0 = rv r (r, θ), f ( re iθ) = e iθ (u r (r, θ) + iv r (r, θ)) = e iθ 1 r = 1 re iθ, f (z) = 1, for all z D. z ( ) Log (1 + i) = Log e i π 4 = ln + i π 4 = ln + iπ 4. log z = Log z + inπ, n Z log (1 + i) = Log (1 + i) + inπ, n Z = ln ( π ) + i 4 + nπ, n Z. (c) As we learned in the theorem of Sec. 7, if a function is analytic on a domain then the comonents are harmonic functions on that domain. Therefore, since Log z is harmonic on the domain D, then its comonents u (x, y) = ln x + iy, v (x, y) = Arg z are harmonic on D. 8