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WA 2: Solutions
Problem 2. Discuss the continuity at z = 0 of the functions defined as (a) Re z 1+|z| ; (b)^
Re z z ; (c)^
(Re z^2 )^2 z^2 for all^ z^6 = 0, and equal 0 at^ z^ = 0. Solution. (a) lim z→ 0 f (z) = lim z→ 0
Re z 1+|z| =^ √ lim x^2 +y^2 → 0
x 1+
x^2 +y^2 = 0 = f (0), thus f
is continuous at z = 0. (b) lim x→ 0 f (x) = lim x→ 0
x x = 1^6 = 0 =^ f^ (0),^ thus^ f^ is discontinuous at^ z^ = 0. (c) lim z→ 0 |f (z)| = lim z→ 0
|Re z^2 |^2 |z|^2 ≤^ zlim→ 0
|z^2 |^2 |z|^2 = lim z→ 0 |z|
(^2) = 0, therefore lim z→ 0 f (z) = 0 =
f (0), i.e., f is continuous at z = 0.
Problem 3. Verify that the function
f (z) =
{ (^) z 2 z when^ z^6 = 0, 0 when z = 0
satisfies the Cauchy–Riemann equations at z = 0, and that f (z) is not differentiable at z = 0.
Solution. Since z
2 z =^
z^3 zz =^
(x−iy)^3 x^2 +y^2 =^
x^3 − 3 ix^2 y− 3 xy^2 +iy^3 x^2 +y^2 ,^ we have^ f^ (z) = u(x, y) + iv(x, y), where
u = x^3 − 3 xy^2 x^2 + y^2
, v = y^3 − 3 x^2 y x^2 + y^2
if x^2 + y^2 > 0, and u = v = 0 if x = y = 0. Then
ux(0, 0) =
d dx [x] = 1 =
d dy [y] = vy (0, 0),
uy (0, 0) =
d dy
[x] = 0 = −
d dx
[y] = −vx(0, 0).
Thus the CR equations at z = 0 are satisfied. However, f is not differentiable at z = 0:
lim z→ 0
f (z) − f (0) z
= lim z→ 0
f (z) z
If z = (1 + i)x then
lim x→ 0
f (x + ix) x + ix
= lim x→ 0
(x − ix)^3 2 x^2 (x + ix)
(1 − i)^3 2(1 + i)
however if z = x then
lim x→ 0
f (x) x = lim x→ 0
x^2 x^2
Conclusion: the existence of partial derivatives at the point, and even the validity of CR at the point does not guarantee the differentiability at this point.
Problem 4. Prove that: (i) If f = u + iv is analytic and satisfies u^2 = 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. 1
2
Solution. (i) By CR,
ux = vy =
∂y
(u^2 ) = 2uuy , uy = −vx = −
∂x
(u^2 ) = − 2 uux.
Then uy = −(2u)^2 uy , i.e., (1 + 4u^2 )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 ≡ C^2 , thus f ≡ C + iC^2. (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 6. Check that for any integer m, the functions u(reiθ^ ) = rm^ cos(mθ) and v(reiθ^ ) = rm^ sin(mθ) satisfy the Cauchy–Riemann equations, using (a) the polar form of these equations; (b) the standard form of these equations; (c) the fact that for a function f = u + iv which is differentiable the Cauchy–Riemann equations for u and v are satisfied.
Solution. (a) If z 6 = 0 then we have ur = mrm−^1 cos(mθ), uθ = −mrm^ sin(mθ), vr = mrm−^1 sin(mθ), vθ = mrm^ cos(mθ). Thus, rur = vθ , uθ = −rvr , i.e., the Cauchy–Riemann equations in polar form are satisfied. (b) First observe that f (z) = u(r, θ) + iv(r, θ) = rmeimθ^ = zm^ = (x + iy)m. If m = 0 then f (z) ≡ 1, and obviously ux = vx = uy = vy ≡ 0, and the CR equations are satisfied. If m > 0 then by the binomial formula,
f (z) = (x + iy)m^ =
∑^ m
k=
m! k!(m − k)!
xm−kikyk,
and then
u =
0 ≤k≤m,even
m! k!(m − k)!
xm−kikyk^ =
[∑ m 2 ]
j=
(−1)j^
m! (2j)!(m − 2 j)!
xm−^2 j^ y^2 j^ ,
v =
0 ≤k≤m,odd
m! k!(m − k)!
xm−kikyk
[ m∑− 2 1 ]
j=
(−1)j^ m! (2j + 1)!(m − 2 j − 1)!
xm−^2 j−^1 y^2 j+1.
Here [r] denotes the integer part of r ∈ R. We have
ux =
[∑ m 2 ]
j=
(−1)j^
m! (2j)!(m − 2 j)!
(m − 2 j)xm−^2 j−^1 y^2 j
[ m∑ 2 − 1 ]
j=
(−1)j^
m! (2j)!(m − 2 j − 1)!
xm−^2 j−^1 y^2 j^ ,
