- #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)##.