f(x + h) f(x) h as representing the slope of a secant line. As h goes to 0, the slope of the secant line approaches the slope of the tangent line.

Derivative of f (z) Dr. E. Jacobs The derivative of a function is defined as a limit:

f0(x) = lim

h→0

f (x + h) − f (x) h

We can visualize the expression f (x+h)−f (x)_h as representing the slope of a 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:

f0(x) = lim h→0 (x + h)3− x3 h = lim h→0 x3+ 3x2h + 3xh2 + h3− x3 h = lim h→0 ¡ 3x2+ 3xh + h2¢ = 3x2

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:

f0(z) = lim

∆z→0

f (z + ∆z) − f (z) ∆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) = z3_{. Find f}0_(z)

f0(z) = lim ∆z→0 (z + ∆z)3 _{− z}3 ∆z = lim ∆z→0 z3 _{+ 3z}2_{∆z + 3z(∆z)}2 _{+ (∆z)}3_{− z}3 ∆z = lim ∆z→0 ¡ 3z2 + 3z∆z + (∆z)2¢ = 3z2

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 dz (z n_{) = nz}n−1 d dz(sin z) = cos z d dz(cos z) = − sin z d dz(sinh z) = cosh z d dz(cosh z) = sinh z d dz (e z_{) = e}z d dz(f (z) + g(z)) = f 0_{(z) + g}0_(z) d dz(f (z) · g(z)) = f (z)g 0_{(z) + f}0_(z)g(z) d dz µ f (z) g(z) ¶ = g(z)f 0_{(z) − f (z)g}0_(z) (g(z))2 d dz(f (g(z)) = f 0_{(g(z)) · g}0_(z)

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) h = limh→0− f (x + h) − f (x) h

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 Cauchy-Riemann 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: f0(z) = lim ∆z→0 f (z + ∆z) − f (z) ∆z = lim ∆z→0 ∆u + i∆v ∆z

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 f0(z) = lim ∆z→0 ∆u + i∆v ∆z = lim∆x→0 ∆u + i∆v ∆x = lim ∆x→0 µ ∆u ∆x + i ∆v ∆x ¶ = ∂u ∂x + i ∂v ∂x

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 f0(z) = lim ∆z→0 ∆u + i∆v ∆z = lim∆y→0 ∆u + i∆v i∆y = lim ∆y→0 µ 1 i ∆u ∆y + ∆v ∆y ¶ = 1 i ∂u ∂y + ∂v ∂y Of course 1_i = 1_i · i_i = _ii2 = −i so,

f0(z) = 1 i ∂u ∂y + ∂v ∂y = −i ∂u ∂y + ∂v ∂y

This leads to two different formulas for the derivative that are supposed to give the same answer.

f0(z) = ∂u ∂x + i ∂v ∂x f0(z) = ∂v ∂y − i ∂u ∂y

As an example, consider the function f (z) = z3_{whose derivative f}0_{(z) = 3z}2

was calculated earlier using limits. This time, let’s try doing the calculation using f0_{(z) =} ∂u

Since z3 _{= (x + iy)}3 _{= x}3 _{− 3xy}2_{+ i(3x}2_{y − y}3_{) it follows that:} u = x3− 3xy2 v = 3x2y − y3 ∂u ∂x = 3x 2_{− 3y}2 ∂v ∂x = 6xy Therefore, the derivative must be:

f0(z) = ∂u ∂x + i

∂v ∂x

= 3(x2− y2) + 6xyi = 3¡x2+ 2xyi − y2¢= 3(x + iy)2 = 3z2 This is the same formula obtained earlier. We could have just as easily used the formula f0_{(z) =} ∂v

∂y − i∂u∂y

f0(z) = ∂v ∂y − i ∂u ∂y = ∂ ∂y ¡ 3x2y − y3¢− i ∂ ∂y ¡ x3− 3xy2¢ = 3x2− 3y2− i (−6xy) = 3¡x2 + 2xy − y2¢ = 3(x + iy)2 = 3z2

Since ∂u_∂x + i∂v_∂x = ∂v_∂y − i∂u_∂y then we may compare the real parts of both sides of the equation and the imaginary parts of both sides to obtain:

∂u ∂x = ∂v ∂y ∂v ∂x = − ∂u ∂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) =}

x2 + y2 and v(x, y) = 0. ∂u_∂x = 2x will only equal ∂v_∂y = 0 at x = 0. ∂v_∂x = 0 will only equal −∂u_∂y = −2y at y = 0. Consequently, the function f (z) = |z|2

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 = exeiy = ex(cos y + i sin y) = excos y + iexsin y u = excos y v = exsin y

∂u

∂x = exsin y is the same as ∂v∂y at all points. Also, ∂x∂v = exsin y is the same

as −∂u_∂y = −(−ex_{sin y) = e}x_{sin y at all points. Therefore, the derivative}

exists at all points and is given by: f0(z) = ∂u

∂x + i ∂v ∂x = e

x_{cos y + ie}x_{sin y = e}x_{(cos y + i sin y) = e}x+iy _{= e}z

Definition A function f (z) is said to be analytic at a point z0 if the

deriva-tive 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.