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Patrick Corn Math 185, 6/28/ Solution Set 2
Problem 1: Let a and b be two points in U. Let γ be a differentiable path from a to b. Consider h = f ◦ γ : [0, 1] → C. Let γ = α + iβ, and let f = u + iv, and let h = h 1 + ih 2. Then h 1 and h 2 are differentiable (in the Math 1A sense!) with derivatives given by
h′ 1 (t 0 ) = ux(x 0 , y 0 )α′(t 0 ) + uy (x 0 , y 0 )β′(t 0 ) h′ 2 (t 0 ) = vx(x 0 , y 0 )α′(t 0 ) + vy (x 0 , y 0 )β′(t 0 )
where x 0 + iy 0 = γ(t 0 ). This is just the chain rule. The point is, however, that the partial derivatives of u and v are all zero because f ′^ is identically zero. This implies that h′ 1 and h′ 2 are identically zero. Now this, in turn, implies that h 1 and h 2 are constant (you may even remember the proof: good old Mean Value Theorem). So h is constant. Then h(0) = h(1), so f (a) = f (b). This holds for any two points in U , so we’re done. As for the extra credit, ask me if you’re interested. The hard part is the word “differentiable.” Problem 2: On the real and imaginary axes, f (z) is identically zero. A moment’s thought, then, will convince you that ux = uy = vx = vy = 0 at (0, 0). But
lim x→ 0
f (x + ix) x + ix
= lim x→ 0
|x| x(1 + i)
(where now x ∈ R) doesn’t even exist. (It’s approaching different values from the left and right). So the derivative of f at 0 cannot exist.
Problem 3: Think of x = r cos θ and y = r sin θ as intermediate variables. So then u and v are functions of x and y, which are in turn functions of r and θ. This is a perfect situation for the chain rule. So:
ur = uxxr + uy yr = ux cos θ + uy sin θ vθ = vxxθ + vy yθ = −rvx sin θ + rvy cos θ uθ = uxxθ + uy yθ = −rux sin θ + ruy cos θ vr = vxxr + vy yr = vx cos θ + vy sin θ
and then it is easy to see that ux = vy and uy = −vx imply rur = vθ and uθ = −rvr. (There is a minor subtlety: the latter two equations also imply the standard Cauchy-Riemann equations, and to completely rigorously answer the question, you should also show this. But I’m not too worried about this.)
Problem 4: (a) This should follow directly from problem 1, because the line segment between any two points in D is certainly a differentiable path (at least when we give it the right parame- terization).
(b) The condition implies that v = 0, so that vx = vy = 0. Then the Cauchy-Riemann equations give ux = uy = 0 also. So f ′^ is identically zero and we’ve reduced to part (a).
(c) I was wrong about polar Cauchy-Riemann–it doesn’t apply here at all. Let’s do it right. The condition is equivalent to saying that u^2 + v^2 is constant in D. We can assume that this constant is nonzero, because otherwise there is nothing to prove–if |f | = 0 everywhere, then f is identically
zero. Taking partials, we get
2 uux + 2vvx = 0 2 uuy + 2vvy = 0
and then we can use the Cauchy-Riemann equations. We get ( 2 u − 2 v 2 v 2 u
ux uy
The determinant of the above matrix is 4(u^2 + v^2 ), which is a (nonzero) constant! Then we can multiply by the inverse of the first matrix to get ux = uy = 0. This implies that vx = vy = 0 by Cauchy-Riemann, and we are done, just as in part (b).
(d) Here we have that v/u is constant. (If u or v vanishes somewhere, then they vanish every- where, by the requirement, and then we can proceed as in part (b). So we can assume that v and u are everywhere nonzero.) Taking partials, and multiplying out the denominators:
uvx − vux = 0 uvy − vuy = 0
After using Cauchy-Riemann, we get ( u −v v u
vx vy
The determinant never vanishes (since we’re already assuming u and v never vanish), so we can conclude that vx = vy = 0 and proceed as in part (b).
Problem 5: You should not just be showing that the Cauchy-Riemann equations hold, because of the asymmetry discussed in the text–we do not yet know that holomorphic functions have continuous first partial derivatives (though it is true). So the Cauchy-Riemann equations alone are not enough. Instead, we just use the limit definitions. We want to show that the limit of the difference quotient for g at z 0 exists, for z 0 ∈ G. So:
lim z→z 0
g(z) − g(z 0 ) z − z 0
= lim z→z 0
f (z) − f (z 0 ) z − z 0
= lim w→z 0
f (w) − f (z 0 ) w − z 0
= f ′(z 0 ).
(Here w = z.) So the limit exists and equals f ′(z 0 ). Problem 6: Well, (u^2 )xx = (2uux)x = 2(uuxx + (ux)^2 ). Similarly for uyy. So, using that u^2 and u are harmonic, we get
0 = uxx + uyy = 2u(uxx + uyy ) + 2((ux)^2 + (uy )^2 ) = 2((ux)^2 + (uy)^2 ),
and so ux = uy = 0 everywhere, so that u is a constant.
Problem 7: Let us first show that if φ preserves the upper half plane, then it is induced by a matrix with real entries and unit determinant. Well, in this case, the real axis maps to itself. So let z 1 , z 2 , z 3 ∈ C be the three points on the extended real axis going to 0, 1 , ∞, respectively. From Step 2 of our proof, φ is induced by a matrix with real entries. (Just look at the formulas–they’re also in the book on p. 27.)
inequality, we would have
ǫ = |f (a)| = |f (a) − f (z) + f (z)| ≤ |f (a) − f (z)| + |f (z)| < ǫ + 0,
which is absurd. So f (z) is nonzero for all z ∈ N , as required.
