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1). use def. delta epsilon proof to prove lim(z goes to z0) Re(z)=Re(z0)

This is what I did |Re(z)-Re(z0)| = |x-x0| < epsilon then |z-z0|=|x-iy-x0-iy0|=|x-x0+i(y-y0)|<=|x-x0|+|y-y0|=epsilon + |y-y0| = delta

My question is doesn't this delta have to be a function of epsilon so is it alright to write it with this |y-y0| value in there? I don't see how to do it otherwise

2). lim(z goes to z0) conjugate of z = conjugate of z0

the way I see this problem as since the conjugate is just a reflection over the real axis then the delta value would be the same as the epsilon value because the points are the same distance away |z-z0|=|conjugate(z)-conjugate(z0)|

but how exactly would you go about writing this as a formal proof?

3). One interpretation of a function w=f(z)=u(x,y)+iv(x,y) is that of a vector field in the domain of definition of f. The function assigns a vector w, with components u(x,y) and v(x,y), to each point z at which it is defined. Indicate graphically the vector fields represented by

a) w=iz

b). w=z/|z|

for a) I have u(x,y) = -y and v(x,y)=x and for b) I have

u(x,y)=x/sqrt(x^2 + y^2) and v(x,y)=y/sqrt(x^2 + y^2)

my question is how do you show graphically this vector field which I think is just the image without a domain?

thanks for any help