Complex differentiability of sin |z|, cos |z| and |z|
1. sin |z| at 0. CW4 Exercise-3-b-i asks about the differentiability of sin |z|, z ∈ C. Let
us check the differentiability at point 0. Note that since |z| ∈ R+ , the function sin |z| is
fully determined by the values of the sine function on reals.
The Taylor expansion at 0 with two terms is
sin x = 0 + x + o(x),
where o(·) denotes a function of x of ‘order smaller than x as x → 0’, meaning that
= 0. Substituting this in the divided difference underlying the definition of
limx→0 o(x)
x
derivative at 0
|z| + o(|z|)
sin |z| − 0
=
z−0
z
and noting that
o(|z|) o(|z|) =0
lim
lim = z→0
z→0 z
|z| we are reduced to inspecting the ratio
|z|
.
z
(1)
If you take z ∈ R+ and let it converge to 0 (notation z → 0+ ), then (1) will converge
to 1, because in that case |z|/z = z/z. Likewise, for real z → 0− the limit of (1) is −1
(why?). Hence z 7→ |z| is not differentiable at 0 as function of z ∈ R on the real line,
whence nondifferentiability of sin |z| at 0. To see the ‘complex dimension’ of the picture,
consider the polar form z = reiθ and write (1) as
r
= e−iθ .
iθ
re
Keeping θ fixed and sending r = |z| → 0 we see that (1) converges to a value that depends
on the direction θ.
Exercise: construct a sequence z1 , z2 , . . . of complex numbers such that
zn
|zn |
oscillates from i to −i and back infinitely many times as n varies.
1
2. cos |z| at 0. What about differentiability of cos |z| at 0? We need to look at the
divided difference
cos |z| − 1
.
z−0
Plugging the Taylor expansion cos |z| = 1 + 0 · |z| + o(|z|) and passing to limit as z → 0
it is seen that cos |z| is diff-le at 0 with
d(cos |z|) =0
dz
z=0
This example responds to the CW5 Exercise 1-b.
The function z → |z| itself is not diff-le at 0, while sin z, cos z are. Why then this
difference between sin |z| and cos |z|? An intuitive explanation lies in the chain rule for
composite functions:
(f (g(z))0 = f 0 (g(z))g 0 (z).
(2)
In the situation in focus each of f (z) = sin z and f (z) = cos z is diff-le while g(z) = |z|
not. However, in the cosine case the non-diff-ty of |z| is annihilated by the flatness of
cosine at 0, to the extent that the composition is diff-le: then the right-hand side of (2)
reads as ‘0 times indefinite = 0’. In contrast to that, in the sine case we only have ‘1
times indefinite = indefinite’. In the next section we make this precise by estimating the
contribution of ‘indefinite’.
3. A more global picture of diff-ty for sin |z|, cos |z|. Exercise 3-a (CW 4) tells
us that entire real-valued function must be constant. The very same argument using
CR-equations applies to show that a holomorphic real-valued function must be constant.
It follows that none of |z|, sin |z|, cos |z| isp
holomorhic. This looks truly surprising if you
think in terms of real analysis: e.g. |z| = x2 + y 2 is smooth (has any number of partial
derivatives) everywhere except z = 0.
The nonholomophicity does not preclude, however, diff-ty at some points. We have
seen that cos |z| is diff-le at point 0. Are there any other such points? Yes, these are
many! And they are the points of the concentric circles with radia π, 2π, 3π, . . . :
{z ∈ C : |z|/π ∈ Z+ }.
To see this, we wish to estimate the divided difference
cos |z| − cos |z0 |
z − z0
as z → z0 for fixed z0 ∈ C. Start with Taylor expansion of cos in the variable |z|
cos |z| − cos |z0 | = (cos0 |z0 |)(|z| − |z0 |) + o(|z| − |z0 |)
(where cos0 u = − sin u, of course).
Then recall the triangle inequality to get
|z| − |z0 | ≤ |z − z0 |
2
and observe that it implies o(|z| − |z0 |) = o(|z − z0 |) (think carefully of the o-notation!).
From this and with another application of the triangle inequality we estimate
cos |z| − cos |z0 | ≤ |(cos0 |z0 |)| · |z − z0 | + o(|z − z0 |).
Diving by |z − z0 | yields
cos |z| − cos |z0 | ≤ |(cos0 |z0 |)| + o(|z − z0 |) .
z − z0
|z − z0 |
Sending z → z0 (or, what is the same, varying z in such a way that |z − z0 | → 0), the
second term in the right-hand side converges to 0. Thus if cos0 |z0 | = 0 then the divided
difference goes to zero, hence in that case cos |z| is diff-le at z0 with derivative 0. This
is the case precisely when − sin |z0 | = 0, that is |z0 |/π ∈ Z+ (equivalently z0 = πkeiθ for
some nonnegative integer k and some θ ∈ [0, 2π)).
If − sin |z0 | =
6 0 then an argument similar to that in section 1 shows that cos |z| is not
diff-le at z0 .
Exercise. Find out where sin |z| is diff-le.
Exercise. Find a function f which is differentiable at 0, not differentiable at any other
point of a disc centered at 0, and has f 0 (0) 6= 0.
4. Nondifferentiability of |z|. Finally, we check that |z| is nowhere differentiable. We
will take fixed nonzero z0 = r0 eiθ0 and vary z = reiθ .
First look at the ‘radial’ divided differences for z = reiθ0 , with z running on the ray
connecting z and 0:
r − r0
|z| − |z0 |
= iθ0
= e−iθ0
z − z0
re − r0 eiθ0
(recall from section 1 the cases θ0 = 0 and θ0 = π).
Thus if the derivative of z 7→ |z| at r0 eiθ0 exists, it must be equal e−iθ0 . There is no
problem so far, because θ0 is fixed. But now let z running on the same circle as z0 , by
letting z = r0 eiθ and varying only θ:
r0 − r0
0
|z| − |z0 |
=
= iθ
= 0.
iθ
iθ
0
z − z0
r0 e − r0 e
e − eiθ0
We see that the divided difference is exactly eiθ0 if z varies radially, and 0 if z is
incident to the same circle as z0 . Since eiθ0 6= 0 for all θ0 ∈ [0, 2π), the function z 7→ |z|
is not diff-le in every point z 6= 0. The case z = 0 was treated in section 1. Hence |z| is
nowhere diff-le in C.
3