WA 3: Solutions - Drexel University

WA 3: Solutions

Problem 1. Prove that: (i) If f = u + iv is analytic and satisfies u2 = v in a domain D, then f is a constant. (ii) If f is a real-valued analytic function in a domain D, then f is a constant.

Solution. (i) By CR,

ux

=

vy

=

(u2) y

=

2uuy ,

uy

=

-vx

=

- (u2) x

=

-2uux.

Then uy = -(2u)2uy, i.e., (1 + 4u2)uy = 0 which is possible only if uy = 0. Then we have also ux = vx = vy = 0, which implies that u C R, and v C2, thus f C + iC2.

(ii) Since v 0, we have vx = vy 0. Then by CR, ux = 0, uy = 0, and thus

u C R and then f C.

Problem 2. Find all the points where the function f (z) = yx2 + i(xy2 - x) is (a) differentiable; (b) analytic. Justify your answers.

Solution. (a) We have u = yx2, v = xy2 - x. The equality ux = 2xy = vy is always true, and the equality uy = -vx is true when x2 = 1 - y2. Therefore, the CR equations are satisfied on the circle x2 + y2 = 1 (or, equivalently, on |z| = 1). Since all the first order partial derivatives of u and v are continuous everywhere, this guarantees that f is differentiable on the circle |z| = 1.

(b) Since the differentiability set for f (the unit circle |z| = 1) has empty interior, this function is analytic nowhere. (Analyticity at some point = Differentiability in some neighborhood of the point).

Problem 3. Prove the following version of the reflection principle. Suppose that a function f is analytic in some domain D which is symmetric about the imaginary axis and contains a segment I of the imaginary axis. Then f (z) = -f (-z) for every z D if and only if f (iy) is purely imaginary for every iy I.

Note: The fact that the first order partial derivatives of the real and imaginary parts of the analytic function f are continuous in D can be used without proof. (Where you are using this fact?)

Solution. If f (z) = -f (-z) for every z D, then in particular f (iy) =

-f (-iy)

=

-f (iy)

for

every

iy

I,

which

means

that

Ref (iy)

=

f (iy)+f (iy) 2

=

0

for every iy I.

Conversely, suppose that f (iy) is purely imaginary for every iy I. Define

the function F (z) = -f (-z) = -u(-x, y) + iv(-x, y), where u(x, y) = Ref (z),

v(x, y) = Imf (z). Let U (x, y) = ReF (z) = -u(-x, y) and V (x, y) = ImF (z) =

v(-x, y). Then Ux(x, y) = ux(-x, y) = vy(-x, y) = Vy(x, y) and Uy(x, y) = -uy(-x, y) = vx(-x, y) = -Vx(x, y) by the CR for u(x, y) and v(x, y), i.e., the CR equations are satisfied for U (x, y) and V (x, y). Since the first order partial

derivatives of u and v are continuous, so are the first order partial derivatives of

U and V . Together with CR, this implies analyticity of F (here it is -- we used

the fact!). Now, since F (iy) = -f (iy) = f (iy) for every iy I, according to our

1

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assumption, and since I has accumulation points in I (in fact, all its points are accumulation points), the uniqueness theorem implies that F (z) = f (z) for all z D, i.e., -f (-z) = f (z) in D, which is equivalent to f (z) = -f (-z) for every z D.

Problem 4. Let the function f (z) = u(r, ) + iv(r, ) be analytic in a domain D that does not include the origin. Using the Cauchy?Riemann equations in polar coordinates and assuming continuity or partial derivatives, show that throughout D the function u(r, ) satisfies the partial differential equation

r2urr(r, ) + rur(r, ) + u(r, ) = 0,

which is the polar form of Laplace's equation. Show that the same is true of the function v(r, ).

Solution. The CR equations in polar coordinates are rur = v, rvr = -u. Differentiating both equations with respect to r, we obtain

(1)

rurr + ur = vr and rvrr + vr = -ur.

On the other hand differentiating the CR equations with respect to , we obtain

(2)

rur = v and rvr = -u.

Assuming the continuity of second order partial derivatives, we have

(3)

ur = ur and vr = vr.

Therefore, multiplying the equalities in (1) by r and using (2) and (3), we obtain

0 = r2urr + rur - rvr = r2urr + rur - rvr = r2urr + rur + u

and 0 = r2vrr + rvr + rur = r2vrr + rvr + rur = r2vrr + rvr + v,

as required.

Problem 5. Show that (a) the set of values of log(i1/2) is the same as the set of values of (1/2) log i;

(b) the set of values of log(i2) is not the same as the set of values of 2 log i. Does this contradict to the identity log(z1z2) = log(z1) + log(z2)?

Solution.

(a)

i1/2

=

ei(

4

+k)

,

k

=

0, 1.

Then

log(i1/2)

=

ln

1+i(

4

+k+2n)

=

i(

4

+ m),

where n Z, m = k + 2n,

thus

m is

any integer.

On the other hand,

log i

=

ln 1

+

i(

2

+

2n)

=

i(

2

+

2n),

n

Z,

and

(1/2) log i

=

i(

4

+

n),

n

Z,

i.e., the set of values of log(i1/2) is the same as the set of values of (1/2) log i.

(b) We have log(i2) = log(-1) = i(1 + 2n), n Z; 2 log i = 2i

2

+

2k

=

i(1 + 4k), k Z. Thus, log(i2) = 2 log i as two sets. This does not contradicts to

the identity log(z1z2) = log(z1) + log(z2) because

log i + log i = i

+ 2k 2

+i

+ 2n 2

= i(1 + 2(k + n)), k, n Z,

and thus log(i2) = log i + log i as two sets. This example shows that, in general,

2 log z = log z + log z as two sets.

Problem 6. (a) Use the definition of zc to show that (-1 + 3i)3/2 = ?2 2;

(b) Show that the result in part (a) could have been obtained by writing (-1 + 3i)3/2 = [(-1 + 3i)1/2]3 and first finding the square roots of -1 + 3i;

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(c) Show that the result in part (a) could have been obtained by writing (-1 + 3i)3/2 = [(-1 + 3i)3]1/2 and first cubing -1 + 3i.

Solution.

(a) (-1+ 3i)3/2

=

e3 2

log(-1+ 3i)

=

e3 2

(ln

2+i(

2 3

+2n))

=

2 2ei(3n+1),

n Z, which is 2 2 for odd n and -2 2 for even n.

(b)

(-1

+ 3i)3/2

=

[(-1

+

3i)1/2]3

=

[ 2ei(

3

+k)]3

=

2 2ei(1+3k),

k

=

0,

1,

which is -2 2 for k = 0 and 2 2 for k = 1.

(c) (-1 + 3i)3/2 = [(-1+ 3i)3]1/2 = [-1 +3 3i + 9 - 3 3i]1/2 = 81/2 =

2 2eik, k = 0, 1, which is 2 2 for k = 0 and -2 2 for k = 1.

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