About the notation of derivative used in Cauchy-Riemann equations

Question

This is going to be a very basic ( and dumb ) question but please bear with me, I have a lot of problems in notation whenever I start off with new math and this is the first time that I am being exposed to more FORMAL ways of expressing mathematical ideas so I am a little lost.

I am a first year engineering undergrad and I wanted to start learning Complex Analysis, after a lot of experimentation I have started learning the subject from the book 'Invitation to Complex Analysis' by R.P Boas, but even though the ideas presented in the initial chapters are quite easy to grasp, I am having a really hard time working with the notation and what they mean in the book.

Here are a couple of things I didn't understand:

1. During the proof for the Cauchy-Riemann equations, why is the derivative of the complex function of complex variable, $f(z)$ at a point in the complex plane ( say $z_{0}$ ) $\frac{\partial f}{\partial x}$ AND $\frac{\partial f}{\partial y}$ ( I think it must have to something with different paths of approach, but our equivalent of $calc-III$ wasn't really rigorous ) , i.e. why is: $$ \frac{\partial f}{\partial x} = f'(z_{0}) \\ -i\frac{\partial f}{\partial y} = f'(z_{0}) $$

2. Say $f$ is a complex function of a complex variable i.e. $f(z)$, and say that I am dealing polar coordinates right now, i.e. $(r,\theta)$ then assuming $1.$ ( the above point ) is true then:

$$ \frac{\partial f}{\partial r} = \frac{\partial f}{\partial \theta}=f'(z_{0}) $$ Further, what would the following mean and how would it relate to the previous notation i.e. $\frac{\partial f}{\partial r}$ and $\frac{\partial f}{\partial \theta}$ ( referring to Cauchy-Riemann equations in Polar coordinates ): $$ \frac{\partial z}{\partial r} \&\frac{\partial z}{\partial \theta} $$

( This is my first math question on stack exchange so I apologize if the formatting is not upto the standards that are expected, I just read a primer on $\LaTeX$ five minutes ago and started typing )

Answer 1

1. By definition, $f(z)=f(Re(z)+iIm(z))$ and $f^{\prime}(z)=\lim_{h\rightarrow 0}\frac{f(z+h)-f(z)}{h}$ where the limit is taken with $h$ being complex number. Write $h=x+iy$ then we both have $\lim_{x\rightarrow 0}\frac{f(z+x+i\times 0)-f(z)}{x+i\times 0}=\lim_{h\rightarrow 0}\frac{f(z+h)-f(z)}{h}$ and also $\lim_{y\rightarrow 0}\frac{f(z+0+i\times y)-f(z)}{0+i\times y}=\lim_{h\rightarrow 0}\frac{f(z+h)-f(z)}{h}$ because in both case you are picking a specific direction for the limit. These equality are only true if $f^{\prime}(z)$ mind you, because that mean the full limit exist, then the limit of every directions exist and are the same. So simplify the above equality to get the result.

2. That form of Cauchy-Riemann is false, you need some factor to account for the rate of movement around the circle and the rotation involved when taking from different direction. This is similar to the above: every directions give the same limit if the full limit exist. Here you need to use the fact that if $z\not=0$ then $f^{\prime}(z)=\lim_{h\rightarrow 1}\frac{f(zh)-f(z)}{z(h-1)}$. This is not directly from definition, but follow from the change of variable $k=z(h-1)$ (which has the inverse change of variable $h=1+\frac{k}{z}$, hence the need for $z\not=0$). Then write $h=r\exp(i\theta)$ and proceed as above: fixing $r=1$ for one direction and fixing $\theta=0$ for another. Once you're done, you should see the correct version of Cauchy-Riemann.

3. Write out the polar form $z=r\exp(i\theta)$ then these are partial derivative of $z$ in term of its arguments.

Answer 2

Suppose $z=x+iy$, then $f(z)=f(x+iy)$.

Now we suppose there exists two real-valued functions $u(x,y), v(x,y)$ such that we can rewrite $f(z)=f(x+iy)=u(x,y)+iv(x,y)$. The condition on these functions is that $u,v$ are (real) differentiable at a point in the complex plane.

Now consider our definition of a derivative of $f(z)$,

$$f'(z_0)=\lim_{h\rightarrow 0} \frac{f(z_0+h) - f(z_0)}{h}$$

Using $z_0=x_0+iy_0$, and considering $h$ to be real we have,

$$f'(z_0)=\lim_{h\rightarrow 0} \left (\frac{u(x_0+h, y_0)-u(x_0,y_0)}{h}+ i\frac{v(x_0+h, y_0) - v(x_0, y_0)}{h} \right )$$

$$f'(z_0) = \frac{\partial u(x_0,y_0)}{\partial x}+i\frac{\partial v(x_0,y_0)}{\partial x}$$

Then doing the same for the case where $k$ is real and hence $ik$ is considered to be purely imaginary, we obtain (using the limit as $ik\rightarrow 0$)

$$f'(z_0) = -i\frac{\partial u(x_0,y_0)}{\partial y}+\frac{\partial v(x_0,y_0)}{\partial y}$$

Now all that is left to be done is to compare the real and imaginary parts of the two expressions.

This leaves the Cauchy-Riemann equations:

Equating reals gives, $$\frac{\partial u(x_0, y_0)}{\partial x}=\frac{\partial v(x_0, y_0)}{\partial{y}},$$

and equating imaginary terms gives,

$$-\frac{\partial u(x_0, y_0)}{\partial y}=\frac{\partial v(x_0, y_0)}{\partial{x}}.$$

Because your $x,y$ are purely real then the vector $(x, y)$ is in the basis $(1, i)$ which is orthogonal. This is why we consider the purely real approach using the limit as $h\rightarrow 0$ and then the purely imaginary approach using the limit as $ik\rightarrow 0$.

Then since we can write $z$ in polar form, $x +iy=r(\cos\theta + i\sin\theta)$ we can substitute this into $f(z)$ and differentiate.

$$x(r,\theta)=r\cos\theta, \quad y(r, \theta)=r\sin\theta$$

$$\Rightarrow \frac{\partial u}{\partial r}=\frac{\partial u}{\partial x}\cos\theta + \frac{\partial u}{\partial y}\sin\theta=\frac{1}{r}\frac{\partial v}{\partial \theta}, $$

$$\text{and}\quad \frac{\partial v}{\partial r}=\frac{\partial v}{\partial x}\cos\theta + \frac{\partial v}{\partial y}\sin\theta=\frac{-1}{r}\frac{\partial u}{\partial \theta} $$