1. CRE form

Hi I got a question, which I think may be a trick question.

"State Cauchy Riemann equations for real valued functions u(x,y), v(x,y).

we know a function is generally of the form
f(z) = u(x,y) + iv(x,y).

but U is the real part and V is the imaginary. so why does the question say "state the CRE for REAL VALUED FUNCTIONS u(x,y), v(x,y)."

2. When we separate f(z) as $u(x,y) + i v(x,y)$, u and v are real valued functions. Notice that v is real valued and iv is imaginary.

So if the book gives $f(z)=u(x,y)+iv(x,y)$, then there's no trick.

3. nice. i always was taught that V is the imaginary.

but ill take it as you say since its easier hehe.

4. Well, I should've been more precise.

v(x,y) is a real function. It's also the imaginary part of f(z).

For example, let z = 3 + 5i. The imaginary part of z is 5, which is a real number.

For example, let $f(z) = z^2$. We can write $f(z)=(x+yi)^2 = x^2 - y^2 + i(2xy)$. So here $u(x,y) = x^2-y^2$ and $v(x,y) = 2xy$. Now you say, is 2xy a real or complex function? (x and y are real values).

5. ah, now that coincides with what i was taught aswell.. great job. thanks

and yes, 2xy is a REAL function, and also the imaginary part of u(x,y)

works well. thanks bro.