Complex Valued Functions - 1 (Differential Calculus)
The main issue with complex numbers is that they cannot be ordered.
These issues mainly effect the following class of functions:
\[f:\Bbb C \rightarrow \Bbb C\]Functions with Real Domain & Complex Range
The special case $f:\Bbb R\rightarrow \Bbb C$ is fairly straightforward since $f(x)$ is two rel valed functions:
\[f_{\text{Re}}:\Bbb R\rightarrow \Bbb R;f_{\text{Im}}:\Bbb R\rightarrow \Bbb R\]So that:
\[f(x)=f_{\text{Re}}(x)+if_{\text{Im}}(x)\]Therefore the first derivative is represented by the following:
\[f'(x)=f'_{\text{Re}}(x)+if'_{\text{Im}}(x)\]Integration would give a similar equation:
\[\int^q_p f(x)dx=\int^q_p f_{\text{Re}}(x)dx+i\int^q_p f_{\text{Im}}(x)dx\]Functions with Complex Domain & Range
When $f:\Bbb C\rightarrow \Bbb C$ more care is needed.
Consider $f(z)$ where $z=p+iq$
We have real valued functions $u(p,q)$ and $v(p,q)$:
\[\begin{aligned} &f(z)=f(p+iq)\\ \equiv& f(p,q)=u(p,q)+iv(p,q) \end{aligned}\]So we have a function of two real variables mapping to two real-valued functions of two real variables.
This suggest applying partial derivatives.
Cauchy-Reimann Conditions
In order for $f’(z)$ to be sensibly defined bia partial derivatives we respect to $\text{Re}(z)$ and $\text{Im}(z)$ of $\text{Re}\left(f(z)\right)$ and $\text{Im}\left(f(z)\right)$ (the 2 variable functions $u(p,q)$ and $v(p,q)$ of the previous slide) there are some important preconditions that $u(p,q)$ and $v(p,q)$ have to satisfy.
There are called the Cauchy-Riemann conditions:
\[\frac{\partial u}{\partial p}=\frac{\partial v}{\partial q}\] \[\frac{\partial u}{\partial q}=-\frac{\partial v}{\partial p}\]These are conditions on functions.
These conditions say that it doesn’t matter that there are many ways of approaching zero as they are all equivalent.
Example
For $f(z)=z^2$ where $z=p+iq$. We have:
\[\begin{aligned} u(p,q)&=\text{Re}\left(f(z)\right)\\ &=\text{Re}\left((p+iq)^2\right)\\ &=p^2-q^2\\ v(p,q)&=\text{Im}\left(f(z)\right)\\ &=\text{Im}\left((p+iq)^2\right)\\ &=2pq \end{aligned}\]Lets apply the first Cauchy-Reimann condition:
\[\begin{aligned} \frac{\partial u}{\partial p}&=2p\\ \frac{\partial v}{\partial q}&=2p \end{aligned}\]These two are the same so we know the first condition is met.
And the second condition:
\[\begin{aligned} \frac{\partial u}{\partial p}&=-2q\\ \frac{\partial v}{\partial q}&=2q \end{aligned}\]This condition is also satisfied.
Notice:
\[\begin{aligned} f'(z)&=2z\\ &=2p+2iq\\ &=\frac{\partial u}{\partial p}+i\frac{\partial v}{\partial p}\\ &=\frac{\partial v}{\partial q}-i\frac{\partial u}{\partial q} \end{aligned}\]The Cauchy-Riemann conditions establish the validity of this rule.
Additional Example Notes
The conjugate function does not satisfy the Cauchy-Riemann conditions. This means that it is not able to be differentiated.
Consider $f(z)=\frac{1}{z}$. From the complex division operation we must rewrite this like so:
\[\frac{1}{z}=\frac{\bar z}{\vert z \vert^2}=\frac{\bar z}{\left(\text{Re}(z)\right)^2+\left(\text{Im}(z)\right)^2}\]We can let $z=p+iq$
\[f(p+iq)=\frac{p-iq}{p^2+q^2}=u(p,q)-iv(p,q)\]Here:
\[\begin{aligned} u(p,q)&=\frac{p}{p^2+q^2}\\ v(p,q)&=\frac{-q}{p^2+q^2} \end{aligned}\]These two functions satisfy the Cauchy-Riemann conditions.