Derivative of f (z)
Dr. E. Jacobs
The derivative of a function is defined as a limit:
f (x + h) − f (x)
h→0
h
f 0 (x) = lim
(x)
We can visualize the expression f (x+h)−f
as representing the slope of a
h
secant line. As h goes to 0, the slope of the secant line approaches the slope
of the tangent line.
You learned in Calculus I that the limit definition is sufficient to calculate
derivatives of elementary functions. For example, to find the derivative of
f (x) = x3 , we perform the following limit calculation:
(x + h)3 − x3
h→0
h
3
x + 3x2 h + 3xh2 + h3 − x3
= lim
h→0
h
¡ 2
¢
= lim 3x + 3xh + h2
f 0 (x) = lim
h→0
2
= 3x
We may apply exactly the same limit process to find the derivative of a
complex valued function. If f (z) is a function of a complex variable then
we define:
f (z + ∆z) − f (z)
f 0 (z) = lim
∆z→0
∆z
Since f (z) is actually a vector field, there is no tangent line interpretation
of the derivative here. However, the limit calculations still work the same
way. For example, suppose f (z) = z 3 . Find f 0 (z)
(z + ∆z)3 − z 3
∆z→0
∆z
3
z + 3z 2 ∆z + 3z(∆z)2 + (∆z)3 − z 3
= lim
∆z→0
∆z
¡ 2
¢
= lim 3z + 3z∆z + (∆z)2
f 0 (z) = lim
∆z→0
2
= 3z
All the familiar derivative formulas are consequences of the limit definition.
Consequently, we can show each of the following familiar derivative formulas
using limits:
d n
(z ) = nz n−1
dz
d
(sin z) = cos z
dz
d
(cos z) = − sin z
dz
d
(sinh z) = cosh z
dz
d
(cosh z) = sinh z
dz
d z
(e ) = ez
dz
d
(f (z) + g(z)) = f 0 (z) + g 0 (z)
dz
d
(f (z) · g(z)) = f (z)g 0 (z) + f 0 (z)g(z)
dz
µ
¶
d f (z)
g(z)f 0 (z) − f (z)g 0 (z)
=
dz g(z)
(g(z))2
d
(f (g(z)) = f 0 (g(z)) · g 0 (z)
dz
When we take h → 0 to find the derivative, it is not supposed to matter
whether h is positive or negative.
lim+
h→0
f (x + h) − f (x)
f (x + h) − f (x)
= lim−
h
h
h→0
For a function of a complex variable f (z), when we find the derivative by
letting ∆z → 0, the direction of ∆z is not supposed to matter.
However, since ∆z is a vector in two dimensions, there are
now many more directions that ∆z can approach 0 from. The condition
that we are supposed to get the same answer no matter what direction
∆z is pointing leads to an important pair of equations called the CauchyRiemann equations. Let us begin by writing both z and f (z) in terms of
their real and imaginary parts.
z = x + iy
f (z) = u(x, y) + iv(x, y)
Both ∆z and f (z + ∆z) − f (z) will have real and imaginary parts.
∆z = ∆x + i∆y
f (z + ∆z) − f (z) = ∆u + i∆v
The derivative can now be written in the following form:
f (z + ∆z) − f (z)
∆z→0
∆z
∆u + i∆v
= lim
∆z→0
∆z
f 0 (z) = lim
Let’s take the case where ∆z approaches 0 in a direction parallel to the real
axis.
∆z = ∆x + i∆y = ∆x + i · 0 = ∆x
∆u + i∆v
∆u + i∆v
= lim
∆x→0
∆z→0
∆x
µ ∆z
¶
∆v
∂v
∆u
∂u
+i
+i
= lim
=
∆x→0 ∆x
∆x
∂x
∂x
f 0 (z) = lim
Now, let’s take the case where ∆z approaches 0 in a direction parallel to
the imaginary axis.
∆z = ∆x + i∆y = 0 + i∆y = i∆y
∆u + i∆v
∆u + i∆v
= lim
∆z→0
∆y→0
∆z
i∆y
µ
¶
1 ∆u ∆v
1 ∂u ∂v
= lim
+
=
+
∆y→0
i ∆y
∆y
i ∂y
∂y
f 0 (z) = lim
Of course
1
i
=
1
i
·
i
i
=
i
i2
= −i so,
f 0 (z) =
1 ∂u ∂v
∂u ∂v
+
= −i
+
i ∂y
∂y
∂y
∂y
This leads to two different formulas for the derivative that are supposed to
give the same answer.
∂u
∂v
f 0 (z) =
+i
∂x
∂x
∂v
∂u
f 0 (z) =
−i
∂y
∂y
As an example, consider the function f (z) = z 3 whose derivative f 0 (z) = 3z 2
was calculated earlier using limits. This time, let’s try doing the calculation
∂v
using f 0 (z) = ∂u
∂x + i ∂x .
Since z 3 = (x + iy)3 = x3 − 3xy 2 + i(3x2 y − y 3 ) it follows that:
u = x3 − 3xy 2
v = 3x2 y − y 3
∂u
= 3x2 − 3y 2
∂x
Therefore, the derivative must be:
∂v
= 6xy
∂x
∂u
∂v
+i
∂x
∂x
¡
¢
2
= 3(x − y 2 ) + 6xyi = 3 x2 + 2xyi − y 2 = 3(x + iy)2 = 3z 2
f 0 (z) =
This is the same formula obtained earlier. We could have just as easily used
∂v
the formula f 0 (z) = ∂y
− i ∂u
∂y
∂v
∂u
−i
∂y
∂y
¢
¢
∂ ¡ 3
∂ ¡ 2
3x y − y 3 − i
x − 3xy 2
=
∂y
∂y
= 3x2 − 3y 2 − i (−6xy)
¡
¢
= 3 x2 + 2xy − y 2
f 0 (z) =
= 3(x + iy)2 = 3z 2
∂v
∂v
∂u
Since ∂u
∂x + i ∂x = ∂y − i ∂y then we may compare the real parts of both
sides of the equation and the imaginary parts of both sides to obtain:
∂u
∂v
=
∂x
∂y
∂v
∂u
=−
∂x
∂y
These are called the Cauchy-Riemann Equations and they are a condition
for differentiability. If the Cauchy-Riemann Equations are not true for a
function f (z) then it’s derivative doesn’t exist.
Example.
Consider the function f (z) = |z|2 . In this case, u(x, y) =
∂v
∂v
x2 + y 2 and v(x, y) = 0. ∂u
∂x = 2x will only equal ∂y = 0 at x = 0. ∂x = 0
2
will only equal − ∂u
∂y = −2y at y = 0. Consequently, the function f (z) = |z|
has a derivative at z = 0 but nowhere else.
Example. Does f (z) = ez have a derivative? If so, what is it?
We first identify u and v and then see if the Cauchy-Riemann equations are
satisfied.
ez = ex+iy = ex eiy = ex (cos y + i sin y) = ex cos y + iex sin y
u = ex cos y
v = ex sin y
∂u
∂x
∂v
∂v
= ex sin y is the same as ∂y
at all points. Also, ∂x
= ex sin y is the same
x
x
as − ∂u
∂y = −(−e sin y) = e sin y at all points. Therefore, the derivative
exists at all points and is given by:
f 0 (z) =
∂v
∂u
+i
= ex cos y + iex sin y = ex (cos y + i sin y) = ex+iy = ez
∂x
∂x
Definition A function f (z) is said to be analytic at a point z0 if the derivative of f (z) exists at z0 as well as in some disk around z0 .
For example, the derivative of |z|2 exists at 0 but nowhere else, so |z|2 is
not analytic at 0. On the other hand, the derivative of ez exists not only
at 0 but in all points in any disk surrounding 0. Therefore ez is analytic at
0. Of course, in the case of ez , the function is analytic at all other points
as well.
In the next sections, you’ll see how the theory of analytic functions and the
Cauchy-Riemann equations have applications to calculations involving fluid
flow.