# Thread: Prove f is analytic in C, and find f'(i)

1. ## Prove f is analytic in C, and find f'(i)

Originally Posted by practice exam
Let $$z=x+iy$$. Prove that the function $f(z)=x^3-3xy^2-3+i(3x^2y-y^3)$ is analytic in $\mathbb{C}$, and find $f'(i)$.
My textbook doesn't cover this, that I'm aware, and I must have missed it in lecture. It's a practice exam question, which means it's meant to be done fairly quickly. But I don't see any easy or quick ways to do this.

Any help would be much appreciated!

2. Originally Posted by hatsoff
My textbook doesn't cover this, that I'm aware, and I must have missed it in lecture. It's a practice exam question, which means it's meant to be done fairly quickly. But I don't see any easy or quick ways to do this.

Any help would be much appreciated!
Since your ' $u$' and ' $v$' are continuously differentiable you can apply the Cauchy-Riemann equations.

3. u and v are continuously differentiable and satisfy the CR equations implies that f is holomorphic, sure. But how do I go from "holomorphic" to "analytic" ?

Presumably I need an actual solution so that I can differentiate and plug in i.

4. Is it a fact that $U_x=V_y~\&~Uy_=-V_x$

5. Originally Posted by hatsoff
u and v are continuously differentiable and satisfy the CR equations implies that f is holomorphic, sure. But how do I go from "holomorphic" to "analytic" ?

Presumably I need an actual solution so that I can differentiate and plug in i.
Well, technically holomorphic and analytic are synonymous...but you may not know that yet (assuming what level you're at). But, I just realized this is much easier since a quick check shows that $f(z)=z^3-3$.

6. Originally Posted by Plato
Is it a fact that $U_x=V_y~\&~Uy_=-V_x$
Well, let's see...

$u_x=3x^2-3y_2=v_y$ and $u_y=-6xy=-v_x$. So I guess I could do

$f'(z)=u_x+iv_x=(3x^2-3y^2)+i(6xy)=3(x^2+2ixy-y^2)=3(x+iy)^2=3z^2$. It follows that $f$ has a Taylor series about any $a\in\mathbb{C}$, and that $f'(i)=-3$.

Is that the easiest way to do this though? I suspect that I knew to jump from $f'(z)=(3x^2-3y^2)+i(6xy)$ to $f'(z)=3z^2$ because I was expecting it based on Drexel28's observation. But on an exam that might not be quite so obvious.

Thanks for the help guys!

7. Originally Posted by hatsoff
u and v are continuously differentiable and satisfy the CR equations implies that f is holomorphic, sure. But how do I go from "holomorphic" to "analytic" ?

Presumably I need an actual solution so that I can differentiate and plug in i.
Are you clear on the definition of "holomophic"? A function is "analytic" (or not) at individual points or on a given set. A function that is "holomorphic" is analytic on the entire complex plane.

Also, it might help to remember why we have the "Cauchy-Riemann" equations.

Suppose f(z)= u(x,y)+ iv(x,y) is to be differentiated at $(x_0, y_0)$. Of course,
$\displaytype\frac{df}{dz}= \lim_{h\to 0} \frac{f(z+h)- f(z)}{h}$
but because the complex plane is two dimensional, for that limit to exist, we must get tthe same limit taking any path toward $(x_0, y_0)$. In particular, if we approach along a line parallel to the real axis, we have
$\frac{df}{dz}= \lim_{h\to 0}\frac{u(x_0+h, y_0)- u(x_0,y_0)}{h}+ i\lim_{h\to 0}\frac{v(x_0+h,y_0)- v(x_0,y_0)}{h}$
$\frac{df}{dz}= \frac{\partial u}{\partial x}+ i \frac{\partial v}{\partial x}$

While if we approach $(x_0, y_0)$ along a path parallel to the imaginary axis
$\frac{df}{dz}= \lim_{h\to 0}\frac{u(x_0, y_0+ h)- u(x_0,y_0)}{hi}+ i\lim_{h\to 0}\frac{v(x_0,y_0+h)- v(x_0,y_0)}{hi}$
$\frac{df}{dz}= -i\frac{\partial u}{\partial y}+ \frac{\partial v}{\partial y}$

In order that f be differentiable at $(x_0, y_0)$, those must be the same so, setting real part equal to real part and imaginary part equal to imaginary part, we have
$\frac{\partial u}{\partial x}= \frac{\partial v}{\partial y}$ and
$\frac{\partial u}{\partial y}= -\frac{\partial v}{\partial x}$

Those also tell us how to calculate $\frac{df}{dz}$- use either
$\frac{df}{dz}= \frac{\partial u}{\partial x}+ i \frac{\partial v}{\partial x}$ or

$\frac{df}{dz}= -i\frac{\partial u}{\partial y}+ \frac{\partial v}{\partial y}$