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!
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!
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'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 .
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