1. differentiation with complex variables

I need to find the values for which the following have derivatives and I need to find the derivatives at those points.

(i) f(z) = x^2 + iy

Then u(x,y) = x^2 and v(x,y) = y.
du/dx = 2x
du/dy = 0
dv/dx = 0
dv/dy = 1
This means 2x = 1 and 0 = 0 and f(z) is only differentiable at x = 1/2.
So it's differentiable at points z(1/2,y) and its derivative is df/dx = 2x = 1 at these points.

(ii) f(z) = xy + xyi

Then u(x,y) = xy and v(x,y) = xy.
du/dx = y
du/dy = x
dv/dx = y
dv/dy = y
This means x = y and x = -y and f(z) is only differentiable along these lines. So it's differentiable at points z(x,x) and z(x,-x) and its derivative is df/dx = y + yi = x + xi and x - xi at these points.

(iii) f(z) = y- 2xy + i(-x + x^2 - y^2).

Then u(x,y) = y- 2xy and v(x,y) = -x + x^2 - y^2
du/dx = -2y
du/dy = 1-2x
dv/dx = -1 + 2x
dv/dy = -2y
This means -2y = -2y and 1-2x = 1-2x and f(z) is differentiable everywhere.
So its differentiable at points z(x,y) and its derivative is df/dx = -2y + i(-1 + 2x) at all points.

Can someone just check that this work is accurate and stated correctly? I'm just learning this material.

2. Originally Posted by PvtBillPilgrim
I need to find the values for which the following have derivatives and I need to find the derivatives at those points.

(i) f(z) = x^2 + iy

Then u(x,y) = x^2 and v(x,y) = y.
du/dx = 2x
du/dy = 0
dv/dx = 0
dv/dy = 1
This means 2x = 1 and 0 = 0 and f(z) is only differentiable at x = 1/2.
So it's differentiable at points z(1/2,y) and its derivative is df/dx = 2x = 1 at these points.
ok

(ii) f(z) = xy + xyi

Then u(x,y) = xy and v(x,y) = xy.
du/dx = y
du/dy = x
dv/dx = y
dv/dy = x
This means x = y and x = -y and f(z) is only differentiable along these lines. So it's differentiable at points z(x,x) and z(x,-x) and its derivative is df/dx = y + yi = x + xi and x - xi at these points.
yes, you made a mistake with dv/dy though, it is x

(iii) f(z) = y- 2xy + i(-x + x^2 - y^2).

Then u(x,y) = y- 2xy and v(x,y) = -x + x^2 - y^2
du/dx = -2y
du/dy = 1-2x
dv/dx = -1 + 2x
dv/dy = -2y
This means -2y = -2y and 1-2x = 1-2x and f(z) is differentiable everywhere.
So its differentiable at points z(x,y) and its derivative is df/dx = -2y + i(-1 + 2x) at all points.

Can someone just check that this work is accurate and stated correctly? I'm just learning this material.