Directional Derivative of Complex Function

Okay. I am starting to understand. Along the line x=x0 means I take a point in the complex plane (which is x0+iy0) and I vary y0, using the real part x0 as a constant. So, the direction of increasing is (0,1), which corresponds to the vector i, which means I am only moving along the imaginary axis in the complex plane. Is that it?Yes, that is correct. Along the line x=x0, you are only varying the imaginary part of the complex number z0, which corresponds to moving along the imaginary axis in the complex plane. And the direction of movement is given by the vector (0,1), or simply i.
  • #1
Bashyboy
1,421
5

Homework Statement


We are given that ##f(z) = u(x,y) + iv(x,y)## and that the function is differentiable at the point ##z_0 = x_0 + iy_0##. We are asked to determine the directional derivative of ##f##

1. along the line ##x=x_0##, and

2. along the line ##y=y_0##.

in terms of ##u## and ##v##.

Homework Equations



Definition of directional derivative: ##\lim\limits_{h \rightarrow 0} \frac{f(z+hw)-f(z)}{h}##

where ##w## is a vector designating the direction along which we are differentiating.

The Attempt at a Solution



I am first going to deal with part 1. The direction along which we are differentiating is ##w = x##. Thus,

##\displaystyle \lim\limits_{h \rightarrow 0} \frac{f(z_0+hw)-f(z_0)}{h} = ##

##\displaystyle \lim\limits_{h \rightarrow 0} \frac{f((x_0+iy_0)+hx_0)-f(z_0)}{h} =##

##\displaystyle \lim\limits_{h \rightarrow 0} \frac{f((x_0+ hx_0) + iy_0)-f(z_0)}{h} =##

##\displaystyle \lim\limits_{h \rightarrow 0} \frac{f(x_0(1+ h) + iy_0)-f(z_0)}{h} =##

##\displaystyle \lim\limits_{h \rightarrow 0} \frac{u(x_0(1+ h), iy_0) + iv(x_0(1+h),y_0 - (u(x_0,y_0) + iv(x_0,y_0))}{h} =## (*)

##\displaystyle \lim\limits_{h \rightarrow 0} \frac{u(x_0(1+ h), iy_0) - u(x_0,y_0) + iv(x_0(1+h),y_0) - iv(x_0,y_0)}{h} =##

##\displaystyle \lim\limits_{h \rightarrow 0} \frac{u(x_0(1+ h), iy_0) - u(x_0,y_0)}{h} + \lim\limits_{h \rightarrow 0} \frac{iv(x_0(1+h),y_0) - iv(x_0,y_0)}{h} =##

##\displaystyle \lim\limits_{h \rightarrow 0} \frac{u(x_0(1+ h), iy_0) - u(x_0,y_0)}{h} + i \lim\limits_{h \rightarrow 0} \frac{v(x_0(1+h),y_0) - v(x_0,y_0)}{h} =##

I have two issues: the first is, did I properly substitute at step (*); second, assuming I did properly substitute, is the last line simply the partial derivatives of u and v, with respect x? The reason I ask is, that I am accustomed to seeing the change of a function, as x changes, represented as ##f(x+\delta x, y) - f(x,y)##.
 
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  • #2
As far as finding directional derivatives is concerned, the fact that these are complex numbers is really irrelevant- you want to treat the real and imaginary parts as independent variables.

I'm not sure what you are doing with the "difference quotients". If you do not know what "u" and "v" are the only thing you can do is write the derivatives of f in terms or [itex]u_x[/itex], [itex]u_y[/itex], [itex]v_x[/itex], and [itex]v_y[/itex].

.
 
  • #3
HallsofIvy said:
I'm not sure what you are doing with the "difference quotients".

The reason why I am using the difference quotient is to calculate the directional derivative, which is the method my book presents, as far as I can understand.

HallsofIvy said:
If you do not know what "u" and "v" are the only thing you can do is write the derivatives of f in terms or uxu_x, uyu_y, vxv_x, and vyv_y.

But I don't know such a thing, that f can be written in terms of the partials of u and v, until I have reached the last line of my work.
 
  • #4
Why are you multiplying ##h## by ##x_0##? ##h## multiplies ##w##, which in the first case is equal to ##w=1+0i##, right?
 
  • #5
vela said:
Why are you multiplying hh by x0x_0? hh multiplies ww, which in the first case is equal to w=1+0iw=1+0i, right?

As far as I understand, the ##w = x + 0i##.
 
  • #6
Sorry, I should have written ##w = x_0 + 0i##.
 
  • #7
Funny. "Along the line x = x0" to me would mean the direction vector is ##(0, i)##
Am I way off? I understand the exercise asks for the derivatives in the point ##z_0## and that's also the way you go to work...
 
  • #8
Bashyboy said:
Sorry, I should have written ##w = x_0 + 0i##.

No: the directional derivative of a function ##f(x,y)## in direction ##p = (u,v)## at the point ##(x_0,y_0)## is defined to be
[tex] D_pf(x_0,y_0) = \lim_{t \to 0} \frac{f(x_0 + t u,y_0 + tv) - f(x_0,y_0)}{t} [/tex]
(see, eg., http://en.wikipedia.org/wiki/Directional_derivative ). Sometimes authors will use the vector of length 1 in direction ##p##, so would define the directional derivative as ##\frac{1}{|p|} D_p f ##, where ##D_p f## is as above.
 
Last edited:
  • #9
BvU said:
Funny. "Along the line x = x0" to me would mean the direction vector is ##(0, i)##
Am I way off? I understand the exercise asks for the derivatives in the point ##z_0## and that's also the way you go to work...
You're right. From his work, I was thinking the OP was looking for the partial derivative in the x-direction, but the first part of the problem wasn't asking for that.
 
  • #10
Okay, well I am quite confused. What do we all agree "along the line x=x_0 means?
 
  • #11
Bashyboy said:
Okay, well I am quite confused. What do we all agree "along the line x=x_0 means?

It is the vertical line through the point ##(x_0,0)##. In the complex plane it is the set of points ##\{ x_0 + i t | t \in \mathbb{R} \}##. Its direction is ##(0,1)## in ##\mathbb{R}^2##, or just ##i## in the complex plane. However, in terms of ##u(x,y), v(x,y)## it should be given as ##(0,1)##.
 
Last edited:

1. What is the definition of a directional derivative of a complex function?

The directional derivative of a complex function is a measure of the rate of change of the function in a specific direction in the complex plane.

2. How is the directional derivative of a complex function calculated?

The directional derivative is calculated using the gradient of the complex function and the unit vector representing the desired direction.

3. What is the significance of the directional derivative in complex analysis?

The directional derivative allows us to understand the behavior of a complex function in a specific direction and can help us find maximum and minimum values of the function.

4. Can the directional derivative of a complex function be negative?

Yes, the directional derivative can be negative if the function is decreasing in the specified direction.

5. How does the directional derivative relate to the Cauchy-Riemann equations?

The Cauchy-Riemann equations are necessary conditions for a function to be differentiable. The directional derivative can be used to check if a complex function satisfies these equations, and therefore, is differentiable.

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