Homework in Real Analysis

Math 220A Complex AnalysisSolutions to Homework #1

Prof: Lei NiTA: Kevin McGown

Conway, Page 4, Problem 2.

Basic geometry in R2 says that we have equality in the expression

|z1 + + zn| |z1|+ + |zn|

if and only if all the zi lie on the same ray with base point at the origin. Thisis true exactly when arg zk = arg z` for all k, `, and since arg zk arg z` =arg(zk/z`), this happens if and only if zk/z` 0 for all k, `.

Conway, Page 6, Problem 5.

For z = cis(2pi/n) it is clear that we have

zn 1 = (z 1)(zn1 + + z + 1) = 0 .

Since n 2, we have 0 < 2pi/n < 2pi and therefore z 6= 1. The result follows.

Conway, Page 44, Problem 19.

We write z = z + iy and f(z) = u(x, y) + iv(x, y). Since f is analytic, we knowthat u and v have continuous partial derivatives and ux = vy, uy = vx. If wedefine f(z) = f(z), then we have f(z) = u(x, y) + iv(x, y) with u(x, y) =u(x,y) and v(x, y) = v(x,y). We compute

ux(x, y) = ux(x,y) = vy(x,y),uy(x, y) = uy(x,y) = vx(x,y),vx(x, y) = vx(x,y),vy(x, y) = vy(x,y),

and therefore ux = vy and u

y = vx. This shows that f is analytic.

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