Homework in Real Analysis
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Math 220A Complex Analysis Solutions to Homework #1 Prof: Lei Ni TA: Kevin McGown Conway, Page 4, Problem 2. Basic geometry in R 2 says that we have equality in the expression |z 1 + ··· + z n |≤|z 1 | + ··· + |z n | if and only if all the z i lie on the same ray with base point at the origin. This is true exactly when arg z k = arg z for all k,, and since arg z k - arg z = arg(z k /z ), this happens if and only if z k /z ≥ 0 for all k,. Conway, Page 6, Problem 5. For z = cis(2π/n) it is clear that we have z n - 1=(z - 1)(z n-1 + ··· + z + 1) = 0 . Since n ≥ 2, we have 0 < 2π/n < 2π and therefore z = 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 know that u and v have continuous partial derivatives and u x = v y , u y = -v x . If we define 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 u * x (x, y)= u x (x, -y)= v y (x, -y), u * y (x, y)= -u y (x, -y)= v x (x, -y), v * x (x, y)= -v x (x, -y), v * y (x, y)= v y (x, -y), and therefore u * x = v * y and u * y = -v * x . This shows that f * is analytic. 1
Transcript of Homework in Real Analysis
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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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