Complex analysis quick problem

[N00813]

Homework Statement

f(z) = u(x, y) + iv(x, y)
where z ≡ x + iy. Let the fluid velocity be V = ∇u. If f(z) is analytic, show that
df/dz = V_x − iV_y

Homework Equations

V_x = du/dx
V_y = idu/dy

The CR equations du/dx = dv/dy, du/dy = -dv/dx.

The Attempt at a Solution

I attempted to use this form for d/dz: d/dz = d/dx - i*d/dy.
This gave me df/dz = 2(V_x - iV_y) though.

 

[BiGyElLoWhAt]

Just to clarify before I try to work this out, V and v are different functions, correct?

 

[HallsofIvy]

I have no idea how you could possibly have got that "2". Please show your work.

 

[N00813]

I have no idea how you could possibly have got that "2". Please show your work.

V = grad(u); so V_x = du/dx and V_y = du/dy

df/dz = (d/dx - id/dy)(u+iv) = du/dx + i dv/dx - i du/dy + dv/dy

Using the CR equations, du/dx = dv/dy and du/dy = -dv/dx.

So df/dz = 2(du/dx) -2i(du/dy) = 2(V_x - iV_y).

 

[N00813]

Just to clarify before I try to work this out, V and v are different functions, correct?

Yes, from the question it appears they are different functions.

 

[pasmith]

Homework Statement

f(z) = u(x, y) + iv(x, y)
where z ≡ x + iy. Let the fluid velocity be V = ∇u. If f(z) is analytic, show that
df/dz = V_x − iV_y

Homework Equations

V_x = du/dx
V_y = du/dy

fixed.

The CR equations du/dx = dv/dy, du/dy = -dv/dx.

The Attempt at a Solution

I attempted to use this form for d/dz: d/dz = d/dx - i*d/dy.

That is not an acceptable choice for $d/dz$.

If $f = u + iv$ is analytic in a neighbourhood of $z$, then by definition the following limit exists, and is independent of how $h \in \mathbb{C}$ approaches zero: $$\frac{df}{dz} = \lim_{h \to 0} \frac{f(z + h) - f(z)}{h} = \lim_{h \to 0} \frac{f(x + \mathrm{Re}(h), y + \mathrm{Im}(h)) - f(x,y)}{h}.$$ Thus you can take $h \in \mathbb{R}$ so that $$\frac{df}{dz} = \frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x},$$ but instead you can take $h \in i\mathbb{R}$ so that $$\frac{df}{dz} = -i\frac{\partial f}{\partial y} = -i\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y}$$ and since these are equal we must have $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}$$ which are the Cauchy-Riemann equations.

One also has $$\frac{df}{dz} = \lim_{h \to 0} \frac{f(x + \mathrm{Re}(h), y + \mathrm{Im}(h)) - f(x,y)}{h}\\ = \lim_{h \to 0} \frac{f(x + \mathrm{Re}(h), y + \mathrm{Im}(h)) - f(x, y + \mathrm{Im}(h))}{h} + \lim_{h \to 0}\frac{f(x, y + \mathrm{Im}(h)) - f(x,y)}{h} \\ = \lim_{h \to 0} \frac{\mathrm{Re}(h)}h \frac{\partial f}{\partial x} + \lim_{h \to 0} \frac{\mathrm{Im}(h)}h \frac{\partial f}{\partial y}.$$

 

[N00813]

fixed.



That is not an acceptable choice for $d/dz$.

If $f = u + iv$ is analytic in a neighbourhood of $z$, then by definition the following limit exists, and is independent of how $h \in \mathbb{C}$ approaches zero: $$\frac{df}{dz} = \lim_{h \to 0} \frac{f(z + h) - f(z)}{h} = \lim_{h \to 0} \frac{f(x + \mathrm{Re}(h), y + \mathrm{Im}(h)) - f(x,y)}{h}.$$ Thus you can take $h \in \mathbb{R}$ so that $$\frac{df}{dz} = \frac{\partial f}{\partial x} = \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x},$$ but instead you can take $h \in i\mathbb{R}$ so that $$\frac{df}{dz} = -i\frac{\partial f}{\partial y} = -i\frac{\partial u}{\partial y} + \frac{\partial v}{\partial y}$$ and since these are equal we must have $$\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}, \qquad \frac{\partial v}{\partial x} = -\frac{\partial u}{\partial y}$$ which are the Cauchy-Riemann equations.

One also has $$\frac{df}{dz} = \lim_{h \to 0} \frac{f(x + \mathrm{Re}(h), y + \mathrm{Im}(h)) - f(x,y)}{h}\\ = \lim_{h \to 0} \frac{f(x + \mathrm{Re}(h), y + \mathrm{Im}(h)) - f(x, y + \mathrm{Im}(h))}{h} + \lim_{h \to 0}\frac{f(x, y + \mathrm{Im}(h)) - f(x,y)}{h} \\ = \lim_{h \to 0} \frac{\mathrm{Re}(h)}h \frac{\partial f}{\partial x} + \lim_{h \to 0} \frac{\mathrm{Im}(h)}h \frac{\partial f}{\partial y}.$$

I looked over my notes again, and apparently d/dz = 1/2 (d/dx - i d/dy).

From here, that would suggest $\lim_{h \to 0} \frac{\mathrm{Re}(h)}h = \frac{1}{2}$ and $\lim_{h \to 0} \frac{\mathrm{Im}(h)}h= \frac{-i}{2}$.

I can't think of how to get there, though.