SOLUTION OF TEST 1 MINGFENG ZHAO September 22, 2012 1. [10 Points] Find the real and imaginary parts of z = 3−i 3+i − . Show all steps. 1+i 1−i Proof. Since we know that z = 3−i 3+i − 1+i 1−i = (3 − i)(1 − i) (3 + i)(1 + i) − (1 + i)(1 − i) (1 − i)(1 + i) = 3 − 3i − i − 1 3 + 3i + i − 1 − 2 2 = 2 − 4i 2 + 4i − 2 2 = 2 − 4i − 2 − 4i 2 = −8i 2 = −4i. Hence we know that Re z = 0, and Im z = −4. 1 2 MINGFENG ZHAO 2. [10 Points] Write the polar representation of the complex number z = 1 − i. Use it to find a simple formula for z 6 . Proof. Since z = 1 − i, then |z| = p √ 1 + (−1)2 = 2 and z is located in the 4th quadrant. Hence we get z = = = = 1−i √ √ 2· 1 1 √ −√ i 2 2 π π 2 · cos − i sin 4 4 1 π 2 2 · e−i 4 . Then we get z 6 = 23 · e−i 6π 4 = 8 · e−i 3π 2 = 8 · e−i 3π 2 +i2π π = 8 · ei 2 = 8i. SOLUTION OF TEST 1 3 3. [10 Points] Find two square roots of z = 1 − i in polar representation. (Can you find all three 3rd roots of z = 1 − i?) Proof. By the proof of the Problem 2, then π 1 z = 1 − i = 2 2 · e−i 4 . Then the two square roots of z are: 1 π z1 = 2 4 · e−i 8 , 1 and z2 = 2 4 · ei − π +2π 4 2 1 = 2 4 · ei 7π 8 . The three 3rd roots of z are: 1 π w1 = 2 6 · e−i 12 , 1 w2 = 2 6 · ei − π +2π 4 3 1 7π = 2 6 · ei 12 , 1 and w3 = 2 6 · ei − π +4π 4 3 1 = 2 6 · ei 15π 12 1 = 2 6 · ei 5π 4 . 4 MINGFENG ZHAO 4. [10 Points] Find all four 4th roots of z = 16. Proof. Since z = 16 = 24 , then all four 4th roots of z are: z1 = 2 and z2 = 2 · ei 2π 4 π = 2 · ei 2 = 2i, z3 = 2 · ei 4π 4 = 2 · eπi = −2, and z4 = 2 · ei 6π 4 =2·e 3π 2 i = −2i. SOLUTION OF TEST 1 5 5. [10 Points] Sketch the set {z : 1 < Re z < 2}. Is it open or closed set? Is it convex? Explain. Proof. The graph of the set {z : 1 < Re z < 2} is: It is open and convex. Since every point in {z : 1 < Re z < 2} is an interior point, then {z : 1 < Re z < 2} is open. For any two points p, q in {z : 1 < Re z < 2}, the line segment connecting p and q lies in the set {z : 1 < Re z < 2}, then {z : 1 < Re z < 2} is path connected, in particular, {z : 1 < Re z < 2} is connected. 6 MINGFENG ZHAO 6. [10 Points] Sketch the set {z : 1 ≤ Im z ≤ 2}. Is it open or closed set? Is it convex? Explain. Proof. The graph of the set {z : 1 ≤ Im z ≤ 2} is: It is closed and convex. Since {z : 1 ≤ Im z ≤ 2} contains its boundary points, then {z : 1 ≤ Im z ≤ 2} is closed. For any two points p, q in {z : 1 ≤ Im z ≤ 2}, the line segment connecting p and q lies in the set {z : 1 ≤ Im z ≤ 2}, then {z : 1 ≤ Im z ≤ 2} is path connected, in particular, {z : 1 ≤ Im z ≤ 2} is connected. SOLUTION OF TEST 1 7 7. [20 Points] Among these four sequences, three are convergent and one is divergent. Identify which are which and explain. an = 1+i n2 + 1 bn = i n+1 cn = 4 − 3i 5 n dn = 1+i 4 − 3i n . Proof. a. an converges to 0, as n → ∞, since √ 1+i 2 = |an | = 2 → 0, n + 1 n2 + 1 as n → ∞. b. bn converges to 0, as n → ∞, since |bn | = 1 i = → 0, n + 1 n + 1 as n → ∞. c. cn diverges, as n → ∞. Otherwise, if cn converges, as n → ∞, than cn is Cauchy, in particular, there exists some N > 0 such that for all n ≥ N , we have √ |cn+1 − cn | < 10 . 10 4 − 3i n 4 − 3i n Since |cn | = = 5 = 1, then we know that 5 √ 4 − 3i n+1 4 − 3i n 4 − 3i −1 − 3i 10 10 |cn+1 − cn | = − − 1 = = > . = 5 5 5 5 5 10 Contradiction. d. dn converges to 0, as n → ∞, since 1+i n |dn | = = 4 − 3i √ !n 2 → 0, 5 as n → ∞. 8 MINGFENG ZHAO 8. [20 Points] Among these four series, two are convergent and two are divergent. Identify which are which and explain. A. Proof. A. ∞ X 1+i 2+1 n n=1 B. ∞ X i n + 1 n=1 C. n ∞ X 4 − 3i n=1 5 D. n ∞ X 1+i . 4 − 3i n=1 ∞ X 1+i is convergent. Since for any n ≥ 1, we have 2+1 n n=1 √ 1+i 2 = n2 + 1 n2 + 1 . √ ∞ X 2 1+i is convergent, by M -test, then is convergent 2 n +1 n2 + 1 n=1 n=1 ∞ ∞ ∞ ∞ X X X X i i i 1 B. is divergent. Otherwise, if is convergent, then −i · = n + 1 n + 1 n + 1 n + 1 n=1 n=1 n=1 n=1 ∞ X 1 diverges. is also convergent, which contradicts with n+1 n=1 n n ∞ X 4 − 3i 4 − 3i C. is divergent. By the result of Problem 7, then does not converge to 5 5 n=1 n ∞ X 4 − 3i is divergent. 0, as n → ∞. By the divergent test, we know that 5 n=1 n ∞ X 1+i D. is convergent. Since for any n ≥ 1, we have 4 − 3i n=1 Since ∞ X ∞ X 1 + i n 1 + i n = = 4 − 3i 4 − 3i n=1 Since √ 2 5 < 1, then ∞ X n=1 √ !n 2 . 5 √ !n n ∞ X 1+i 2 converges. By the M -test, we know that is con5 4 − 3i n=1 vergent. Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT 06269-3009 E-mail address: mingfeng.zhao@uconn.edu