SOLUTIONS MANUAL COMPLEX ANALYSIS- A FIRST COURSE WITH APPLICATION ALL CHAPTERS INCLUDED 2023/2024.

SOLUTIONS MANUAL COMPLEX ANALYSIS- A FIRST COURSE WITH APPLICATION ALL CHAPTERS INCLUDED 2023/2024. IV 1 2 3 4 5 6 7 8 9 10 1 X X X X X X X X X X X 2 X X X X X X X X 3 X X X X X X X 4 X X X X 5 X X X X 6 X X X X 7 X X X X X X X X X X X 8 X X X X X X X X 1 I.1.1 Identify and sketch the set of points satisfying. (a) jz 1 ij = 1 (f) 0 < Im z < (b) 1 < j2z 6j < 2 (g) < Re z < (c) jz 1j 2 + jz + 1j 2 < 8 (h) jRe zj < jzj (d) jz 1j + jz + 1j 2 (i) Re (iz + 2) > 0 (e) jz 1j < jzj (j) jz ij 2 + jz + ij 2 < 2 Solution Let z = x + iy, where x; y 2 R. (a) Circle, centre 1 + i, radius 1. jz 1 ij = 1 , j(x 1) + i (y 1)j = 1 , (x 1)2 + (y 1)2 = 12 (b) Annulus with centre 3, inner radius 1=2, outer radius 1. 1 < j2z 6j < 2 , 1 < 2 jz 3j < 2 , , 1=2 < jz 3j < 1 , (1=2) 2 < (x 3)2 + y 2 < 12 (c) Disk, centre 0, radius p 3. jx + iy 1j 2 + jx + iy + 1j 2 < 8 , , (x 1)2 + y 2 + (x + 1) 2 + y 2 < 8 , x 2 + y 2 < p 3 2 (d) Interval [ 1; 1]. jz 1j + jz + 1j 2 , q (x 1)2 + y 2 2 q (x + 1) 2 + y 2 , , q (x 1)2 + y 2 2 2 q (x + 1) 2 + y 2 2 , , q (x + 1) 2 + y 2 x + 1 , q (x + 1) 2 + y 2 2 (x + 1) 2 , y = 0 Now, take y = 0 in the inequality, and compute the three intervals 2 x < 1; then jx 1j + jx + 1j = (x 1) (x + 1) = 2x 2; 1 x 1 then jx 1j + jx + 1j = (x 1) + (x + 1) = 2 2 x > 1; then jx 1j + jx + 1j = (x 1) + (x + 1) = 2x 2: (e) Half–plane x > 1=2. jz 1j < jzj , jz 1j 2 < jzj 2 , jx + iy 1j 2 < jx + iyj 2 , , (x 1)2 + y 2 < x 2 + y 2 , x > 1=2 (f) Horizontal strip, 0 < y < . (g) Vertical strip, < x < . (h) CnR. jRe zj < jzj , jRe (x + iy)j 2 < jx + iyj 2 , x 2 < x 2 + y 2 , jyj > 0 (i) Half plane y < 2. Re (iz + 2) > 0 , Re (i (x + iy) + 2) > 0 , y + 2 > 0 , y < 2 (j) Empty set. jz ij 2 + jz + ij 2 < 2 , jx + iy ij 2 + jx + iy + ij 2 < 2 , , x 2 + (y 1)2 + x 2 + (y + 1) 2 < 2 , x 2 + y 2 < 0