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• Alfaro Rivera Emerson

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Problems 3.1. Express each of the following complex numbers in the form x+iy : (a). (√2 − i) − i(1 − √2i), (b). (2 − 3i)(−2 + i), (c). (1 − i)(2 − i)(3 − i), (d). 4+3i 3 − 4i , (e). 1 + i i+ i 1−i , (f). 1+2i 3 − 4i + 2−i 5i , (g). (1 + √3 i)−10, (h). (−1 + i)7, (i). (1 − i)4. 3.2. Describe the following loci or regions: (a). |z − z0| = |z − z0|, where Im z0 = 0, (b). |z − z0| = |z + z0|, where Re z0 = 0, (c). |z − z0| = |z − z1|, where z0 = z1, (d). |z − 1| = 1, (e). |z − 2| = 2|z − 2i|, (f).

z − z0

z − z1

= c, where z0 = z1 and c = 1, (g). 0 < Im z < 2π, (h). Re z |z − 1| > 1, Im z < 3, (i). |z − z1| + |z − z2| = 2a, (j). azz + kz + kz + d = 0, k ∈ C, a, d ∈ IR, and |k| 2 > ad. 3.3. Let α, β ∈ C. Prove that |α + β| 2 + |α − β| 2 = 2(|α| 2 + |β| 2), and deduce that |α + α2 − β2| + |α − α2 − β2| = |α + β| + |α − β|. 3.4. Use the properties of conjugates to show that (a). (z)4 = (z4), (b). z1 z2z3

= z1 z2z3

. 3.5. If |z| = 1, then show that

az + b bz + a

=1 16 Lecture 3 for all complex numbers a and b. 3.6. If |z| = 2, use the triangle inequality to show that |Im(1 − z + z2)| ≤ 7 and |z4 − 4z2 + 3| ≥ 3. 3.7. Prove that if |z| = 3, then 5 13 ≤

2z − 1 4 + z2

≤ 7 5 . 3.8. Let z and w be such that zw = 1, |z| ≤ 1, and |w| ≤ 1. Prove that

z−w 1 − zw

≤ 1. Determine when equality holds. 3.9. (a). Prove that z is either real or purely imaginary if and only if (z)2 = z2. (b). Prove that √2|z|≥|Re z| + |Im z|. 3.10. Show that there are complex numbers z satisfying |z−a|+|z+a| = 2|b| if and only if |a|≤|b|. If this condition holds, find the largest and smallest values of |z|. 3.11. Let z1, z2, ··· , zn and w1, w2, ··· , wn be complex numbers. Establish

Lagrange’s identity

n k=1 zkwk

2 = n k=1 |zk| 2 n k=1 |wk| 2

− k
0 (y < 0), then Arg z = π/2 (−π/2). (b). If x > 0, then Arg z = tan−1(y/x) ∈ (−π/2, π/2). (c). If x < 0 and y > 0 (y < 0), then Arg z = tan−1(y/x)+π (tan−1(y/x)− π). (d). Arg (z1z2) = Arg z1 + Arg z2 + 2mπ for some integer m. This m is uniquely chosen so that the LHS ∈ (−π, π]. In particular, let z1 = −1, z2 = −1, so that Arg z1 = Arg z2 = π and Arg (z1z2) = Arg(1) = 0. Thus the relation holds with m = −1. (e). Arg(z1/z2) = Arg z1 − Arg z2 + 2mπ for some integer m. This m is

uniquely chosen so that the LHS ∈ (−π, π]. Answers or Hints 3.1. (a). −2i, (b). −1+8i, (c). −10i, (d). i, (e). (1 − i)/2, (f). −2/5, (g). 2−11(−1 + √3i), (h). −8(1 + i), (i). −4. 3.2. (a). Real axis, (b). imaginary axis, (c). perpendicular bisector (passing through the origin) of the line segment joining the points z0 and z1, (d). circle center z = 1, radius 1; i.e., (x − 1)2 + y2 = 1, (e). circle center (−2/3, 8/3), radius √32/3, (f). circle, (g). 0