Calculating Vector Field for f(z)=-iz

[imagemania]

Homework Statement

f(z) = -iz is an analytic function [i being the complex number]
Calculate the vector field v=u(x,y)i -v(x,y)j

Homework Equations

z = x+iy [Im assuming z here is the complex z not a varaible z]

The Attempt at a Solution

I've not learned polar vector calculus yet, so i need to do it via x & y's

z = x+iy
hence f(z) = y-ix
So u(x,y) = y
v(x,y) = -x
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

I don't think this is correct as it is not in the form it says it should be in, the plus and minus are the wrong way...

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

Any help is much appreciated!

 

[mighty2000]

Your attempt at finding the vector field is correct. The minus sign is just a convention and does not affect the actual vector field. Your vector field can be written as v = -x i + y j, which is equivalent to u(x,y)i -v(x,y)j.

To graphically represent this vector field, you can draw arrows at different points on the x and y axes, as well as on the line y=x. The length of the arrow can represent the magnitude of the vector at that point, while the direction of the arrow can represent the direction of the vector. For example, at the point (1,0) on the x-axis, you would draw an arrow pointing to the left with a length of 1, since v(1,0) = -1 i + 0 j. Similarly, at the point (0,1) on the y-axis, you would draw an arrow pointing upwards with a length of 1, since v(0,1) = 0 i + 1 j. And at the point (1,1) on the line y=x, you would draw an arrow pointing to the top left with a length of √2, since v(1,1) = -1 i + 1 j.

I hope this helps. Keep up the good work with your studies!