Derivative of complex-valued function¶

Let $f$ be a function that takes in a real number and outputs a complex number. We call functions like this complex-valued, and we define the limit $\lim_{x \to a} f(x)$ and derivative $f'(x)$ just like the limit and derivative of vector-valued functions, similarly to how we borrowed other operations from vectors.

We assume that the input $x$ is a real number; only the output $f(x)$ is assumed to be a complex number. It's possible to define limits and derivatives so that the input can be a complex number too, but that's beyond the scope of these derivations.

With vectors, we can take limits and derivatives of $\I$ parts and $\J$ parts separately, as derived here and here. For complex numbers, this corresponds to calculating limits or derivatives of real and imaginary parts separately, like this:

For example, the derivative of the complex-valued function $h(x)=\cos(x) + i\sin(x)$ is $$ h'(x) = -\sin(x) + i\cos(x), $$ and that derivative exists for all real numbers $x$ because the derivatives of $\cos$ and $\sin$ exist.

Limit rules¶

We want to figure out whether limit of sum and limit of product work with complex-valued functions. We could modify the derivations of the limit rules to work with complex numbers, but that too is beyond the scope of these derivations. Instead, we do this by calculating limits of real and imaginary parts separately.

Let $f$ and $g$ be complex-valued functions, and suppose that the limits $\lim_{x \to a} f(x)=A+Bi$ and $\lim_{x \to a} g(x)=C+Di$ exist. We write $$ f(x) = f_1(x)+if_2(x) \quad \text{and} \quad g(x) = g_1(x)+ig_2(x), $$ where $f_1$, $f_2$, $g_1$ and $g_2$ are real-valued, and now we have $$ \begin{align} \lim_{x \to a} f_1(x) &= A, \\ \lim_{x \to a} f_2(x) &= B, \\ \lim_{x \to a} g_1(x) &= C, \\ \lim_{x \to a} g_2(x) &= D. \end{align} $$ By expanding and grouping real and imaginary parts together, we get $$ f(x)+g(x) = f_1(x)+g_1(x)+i(f_2(x)+g_2(x)). $$ By calculating limits of real and imaginary parts separately and then applying limit rules for real-valued functions, we get $$ \begin{align} \lim_{x \to a} (f(x)+g(x)) &= \lim_{x \to a}(f_1(x)+g_1(x))+i\lim_{x \to a}(f_2(x)+g_2(x)) \\ &= \left(\lim_{x \to a}f_1(x)+\lim_{x \to a}g_1(x)\right)+i\left(\lim_{x \to a}f_2(x)+\lim_{x \to a}g_2(x)\right) \\ &= A+C+i(B+D). \end{align} $$ On the other hand, we have $$ \begin{align} \left(\lim_{x \to a} f(x)\right)+\left(\lim_{x \to a} g(x)\right) &= (A+Bi)+(C+Di) \\ &= (A+C)+i(B+D), \end{align} $$ so $$ \lim_{x \to a} (f(x)+g(x)) = \left(\lim_{x \to a} f(x)\right)+\left(\lim_{x \to a} g(x)\right). $$

A similar calculation works with multiplication, but we skip that because it's messier. You can do it yourself if you are interested. Let me know if you need help with it.

Derivative rules¶

We will need these derivative rules for complex-valued functions $f$ and $g$:

• If $f(x) = 0$ for all real numbers $x$, then $f(x)$ is a constant.
• Product rule: if $f'(x)$ and $g'(x)$ exists, then $\frac{d}{dx} (f(x)g(x)) = f'(x)g(x)+f(x)g'(x)$.

For the first of these rules, we can write $f(x)$ as $f_1(x) + if_2(x)$, where $f_1$ and $f_2$ are real-valued functions. By calculating derivatives of real and imaginary parts separately, we get $$ f_1'(x) + if_2'(x) = f'(x) = 0 = 0+0i, $$ and comparing real and imaginary parts on both sides gives $f_1'(x)=0$ and $f_2'(x)=0$ for all real numbers $x$. Because $f_1$ and $f_2$ are real-valued, we know from this derivation that they must be constants, so $f$ is also a constant.

To get product rule to work, let's look at its derivation. The derivation contains a drawing that doesn't work with complex-valued functions, but it also has calculations that can be used instead of the drawing, and those calculations work fine with complex numbers. At the end of the derivation, we also used limit of sum and limit of product, and we just derived those for complex-valued functions.

We can similarly get many other derivative rule derivations to work, such as all derivations on this page. However, we cannot get the chain rule $$ \frac{d}{dx} f(g(x)) = f'(g(x))g'(x) $$ to work when $g(x)$ is a complex number, because we haven't defined what the derivative means when the function takes complex numbers as input, and $g(x)$ is the input of $f$ and $f'$ here. This also means that we don't know whether e.g. quotient rule works, because the derivation of the quotient rule uses the chain rule.