Analytic Functions Cauchy & Riemann Equations

[thomas49th]

Hi, this is fairly fundamental and basic, but I cannot seem to make sense of it

I know z = x + iy

and hence a function of this variable would be in the form h = f(z). BUT I do not understand why

f(z) = u(x,y) + iv(x,y)

why so? in z = x + iy, x is the real part and iy is the imaginary part, so why does y have influence in real part and x have influence in the imaginary part in f(z)?

Thanks
Thomas

 

[I like Serena]

Hi thomas49th!

Consider for instance f(z)=z².
What is (x+iy)²?

 

[Mark44]

Mod note: Moved from Precalculus section.

 

[kru_]

I had a bit of trouble grasping this when first exposed to it, as well. What convinced me was to see f(z) as just another complex number, w. And w has its own real and imaginary part, u + iv.

Now if w = f(z) then we can say that w is (the result of) a function of z, and likewise that u and v are functions of z. So w(z) = u(z) + iv(z). But since the value of z is dependent of the values of its real and imaginary parts, x and y, we can think of w(z) = w(z(x,y)) = u(z(x,y)) + iv(z(x,y)). But there is no need for such ugly notation, since the root idea is that w is a function of both x and y. So we simplify to f(z) = w(x,y) = u(x,y) + iv(x,y).

 

[thomas49th]

@ I like... thought someone might say that :)

@ kru_ that has satisfied me for the time being

all is good for now

Thanks :)