# Thread: Derivatives of complex functions

1. ## Derivatives of complex functions

Okay,

I am doing kind of an advanced study thing in one of my calculus courses and I've done pretty well on all the objectives up until this one. I am completely lost when it comes to this.

My objective is to show that for the complex function $f(z)=\overline{z}$, $f'(z)$ does not exist at any point.

So from my brief reading in Schaum's Outline on Advanced Calculus (the only book in the whole library about advanced calc), I know that complex functions can be written as functions of x and y. So, $f(z)=u(x,y)+iv(x,y)$. I know that $f(z)=\overline{z}$ is the conjugate...and that's where I get stuck. I've read about the Cauchy-Riemann equations and how to determine if a function is analytic, and I know that in the complex plane we can approach a particular point from any direction, like in multivariable calc.

So basically, I'm just lost...can anyone try to break this down for me? I'm just not understanding how to work with these complex functions.

Any help is GREATLY appreciated!

2. Originally Posted by I<3Math
Okay,

I am doing kind of an advanced study thing in one of my calculus courses and I've done pretty well on all the objectives up until this one. I am completely lost when it comes to this.

My objective is to show that for the complex function $f(z)=\overline{z}$, $f'(z)$ does not exist at any point.

So from my brief reading in Schaum's Outline on Advanced Calculus (the only book in the whole library about advanced calc), I know that complex functions can be written as functions of x and y. So, $f(z)=u(x,y)+iv(x,y)$. I know that $f(z)=\overline{z}$ is the conjugate...and that's where I get stuck. I've read about the Cauchy-Riemann equations and how to determine if a function is analytic, and I know that in the complex plane we can approach a particular point from any direction, like in multivariable calc.

So basically, I'm just lost...can anyone try to break this down for me? I'm just not understanding how to work with these complex functions.

Any help is GREATLY appreciated!
Notice that $f(x+iy) = \overline{x+iy} = x-iy = u(x,y) + i v(x,y)$ where $u(x,y) = x, v(x,y) = -y$.
To be analytic we need to have by Cauchy-Riemann: $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y} \text{ and }\frac{\partial u}{\partial y} = - \frac{\partial v}{\partial x}$.
But that means, $1 = -1 \text{ and } 0 = -0$.
The first equation is never satisfied.

3. I read through that super quickly so tonight at work I will read over it again and post if I have any further questions. Thank you SOO MUCH! I have a couple others that I have to prove are not differentiable anywhere so just getting a start on one was great!

4. Perfect Hacker,

Thanks so much for the explanation. I had about 15 problems to do and I did all of them tonight. I feel confident in my answers...I am so relieved that you left a really well-written example for me to follow, it made my other ones a breeze, and I feel like I really understand the concept now.

I do have a question though...I've been reading that a complex function is analytic if the Cauchy-Riemann equations are satisfied...exactly what does it mean to be analytic? Does that mean that a derivative exists for a particular function for some set of points? Or is the definition more broad than that?

5. Strictly speaking "analytic" at a point means there is some neighborhood of the point in which the function is infinitely differentiable, its Taylor series about that point exists and converges in that neighborhood and, finally, that the Taylor series converges to that funtion in that neighborhood.

For functions over the real numbers that is a very strong condition. However, in the complex numbers it can be shown that all of that is true as long as the function has a continuous derivative in some neighborhood of the point.

And that entails the Cauchy-Riemann equations. For this problem, one way to do it without using the Cauchy-Riemann equations it to mimic the proof of that.

If f Is differentiable at, say, $z_0$, then the limit defining the derivative, $\lim_{h\rightarrow 0}\frac{f(z_0+h)- f(z_0)}{h}$ must exist. And that means that the limit along any path must be the same.

For functions of real numbers that is only "from the left" or "from the right" and depends only on whether h is positive or negative. But the complex plane is two dimensional so there are many possible "paths". Two are particularly simple:

1) Along a line parallel to the real axis. Then h is a real number. If $z_0= x_0+ iy_0$, then $z_0+ h= x_0+h+ iy_0$. $f(z_0)= \overline{z_0}= x_0- iy_0$ and $f(z_0+ h)= x_0+ h- iy_0$.
So $f(z_0+h)- f(z_0)= x_0+ h- iy_0- (x_0 - iy_0)= h$
We have $\lim_{h\rightarrow 0}\frac{f(z_0+h)- f(z_0)}{h}= \lim_{h\rightarrow 0} \frac{h}{h}= 1$

2) Along a line parallel to the imaginary axis. Then h is an imaginary number but I am going to change notation and write it as "hi" where h is again real.
Now $z_0+ hi= x_0+ i(y_0+ h)$. $f(z_0+h)- f(z_0)= x_0- i(y_0+h)- (x_0- iy_0)= -ih$.
Now $\lim_{z\rightarrow 0}\frac{f(z_0+ih)- f(z_0)}{hi}= \lim_{h\rightarrow 0}\frac{-ih}{hi}= -1$

Since those two limits are not the same the limit itself does not exist and so the function is not differentiable and not analytic.

6. Originally Posted by HallsofIvy
However, in the complex numbers it can be shown that all of that is true as long as the function has a continuous derivative in some neighborhood of the point.
It is even better than that! If $f$ is differenciable on an open disk it automatically is analytic on that disk.
Thus, you do not even need continous derivatives!

7. Gosh guys, thanks a lot for all that info! It makes me love math even more. I can't wait to start my analysis classes after this semester!