1. ## Vector Calculus

I am told that f(z) = -iz (i is the complex number) and i need to calculate the vector field: lets call it n

n = ui - vj [i and j here are the unit vectors]

Where u and v are functions of x and y.

Im not entirely sure where to begin to deduce what u and v are. Any help is much appreciated!

2. Originally Posted by imagemania
I am told that f(z) = -iz (i is the complex number) and i need to calculate the vector field: lets call it n

n = ui - vj [i and j here are the unit vectors]

Where u and v are functions of x and y.

Im not entirely sure where to begin to deduce what u and v are. Any help is much appreciated!
If I understand correctly first you need that $z=x+iy$

This gives

$f(x+iy)=-i(x+iy)=y-ix$

Now we just separate the real and imaginary parts to get

$u(x,y)=y \quad v(x,y)=-x$

3. Thanks for the response, that would make sense, and going on from that:

We know the gradient of a scalar field is a vector field.

∇f(z) = (∂/∂x i + ∂/∂y j + ∂/∂z k)(y-ix)
= -i i + 1 j

Issue is it doesn't match what it says in the question it says n = ui - vj whereas i get the minus's the other way around. Though i could quote it as u = -i v =-1 ??

I also need to draw arrows to represnt this graphically at the y axis, x axis and y=x.

Is there any easy way to see these?

Thanks again!

4. Originally Posted by imagemania
Thanks for the response, that would make sense, and going on from that:

We know the gradient of a scalar field is a vector field.

∇f(z) = (∂/∂x i + ∂/∂y j + ∂/∂z k)(y-ix)
= -i i + 1 j

Issue is it doesn't match what it says in the question it says n = ui - vj whereas i get the minus's the other way around. Though i could quote it as u = -i v =-1 ??

I also need to draw arrows to represnt this graphically at the y axis, x axis and y=x.

Is there any easy way to see these?

Thanks again!
first let me use
$\mathbf{e}_x$ for the basis vector in the x direction and
$\mathbf{e}_y$ for the basis vector in the y direction

When you separate the function into it's real and imaginary parts

$f(x,y)=y\mathbf{e}_x-x\mathbf{e}_y$

This is a map from $f:\mathbb{R}^2\to \mathbb{R}^2$

To take the gradient use the chain rule

$\displaystyle \nable f(z)=\frac{\partial f}{\partial z}\frac{\partial z}{\partial x}\mathbf{e}_x+\frac{\partial f}{\partial z}\frac{\partial z}{\partial y}\mathbf{e}_y$

This gives

$\nabla f =i(1)\mathbf{e}_x+i(i)\mathbf{e}_y=i\mathbf{e}_x-\mathbf{e}_y$

5. Ok im slightly confused. I understand what you've done, instead of using i.j,k typical format, you've changed it to this basis vector. Issue is we haven't been taught this way yet, so Im assuming we're suppose to do it via i,j,k route.

If i take your method how does plotting the field lines work? For instance the x-axis postive to negative?
[As you probably can guess i've only just learnt the basics of vector calculus]

6. Originally Posted by imagemania
Ok im slightly confused. I understand what you've done, instead of using i.j,k typical format, you've changed it to this basis vector. Issue is we haven't been taught this way yet, so Im assuming we're suppose to do it via i,j,k route.

If i take your method how does plotting the field lines work? For instance the x-axis postive to negative?
[As you probably can guess i've only just learnt the basics of vector calculus]
I just used different symbols for the basis vectors the math does not change. I just didn't want to use i to mean two different things.

7. ok so $e_x = \hat{i}$ ?

Now if i wanted to draw the field lines say along the x-axis. I'd let the f(z) = 0:
$i e_x = e_y$
How would i draw the field lines from this information :S

8. I've been relooking at the question and i think i've stated things wrong. Let me re-state the question

"The analytic function f(z) = -iz, calculate the vector field.
Draw small arrows showing the field along the x-axis and y-axis."

Does it want the curl becuase of the arrow requirements?

If so i get $(i + 1) \hat{k}$.
Does it want the gradient of the scalar field (grad) or the curl?

But i still cannot figure out how to draw these arrows.

Hope someone can help

9. Originally Posted by imagemania
I've been relooking at the question and i think i've stated things wrong. Let me re-state the question

"The analytic function f(z) = -iz, calculate the vector field.
Draw small arrows showing the field along the x-axis and y-axis."

Does it want the curl becuase of the arrow requirements?

If so i get $(i + 1) hat{k}$.
Does it want the gradient of the scalar field (grad) or the curl?

But i still cannot figure out how to draw these arrows.

Hope someone can help
I think I see the source of your confusion.

$\mathbb{C}$ is isomorphic to $\mathbb{R}^2$

as Vector Spaces.

So back in post 2 we have that

$f(z)=-iz \iff \vec{F}(x,y)=y\mathbf{e}_x-x\mathbf{e}_{y}$

Now just plug in (x,y) to generate the vector field.

10. Thanks that helps a lot. By the way, did you use maple to generate that graph? If so, what commands do you use to do such a sketch?

Thanks!

11. Originally Posted by imagemania
Thanks that helps a lot. By the way, did you use maple to generate that graph? If so, what commands do you use to do such a sketch?

Thanks!
yes I did.

Code:
with(plots):
fieldplot([y,-x],x=-2..2,y=-2..2,fieldstrength=log);