# What is the derivative of i?

Dec 18, 2014

You can treat $i$ as any constant like $C$. So the derivative of $i$ would be $0$.

However, when dealing with complex numbers, we must be careful with what we can say about functions, derivatives and integrals.

Take a function $f \left(z\right)$, where $z$ is a complex number (that is, $f$ has a complex domain). Then the derivative of $f$ is defined in a similar manner to the real case:

${f}^{p} r i m e \left(z\right) = {\lim}_{h \to 0} \frac{f \left(z + h\right) - f \left(z\right)}{h}$

where $h$ is now a complex number. Seeing as complex numbers can be thought about as lying in a plane, called the complex plane, we have that the result of this limit depends on how we chose to make $h$ go to $0$ (that is, with which path we chose to do so).

In the case of a constant $C$, it's easy to see that it's derivative is $0$ (the proof is analogous to the real case).

As an example, take $f$ to be $f \left(z\right) = \overline{z}$, that is, $f$ takes a complex number $z$ into it's conjugate $\overline{z}$.

Then, the derivative of $f$ is

${f}^{p} r i m e \left(z\right) = {\lim}_{h \to 0} \frac{f \left(z + h\right) - f \left(z\right)}{h} = {\lim}_{h \to 0} \frac{\overline{z + h} - \overline{z}}{h} = {\lim}_{h \to 0} \frac{\overline{h} + \overline{z} - \overline{z}}{h} = {\lim}_{h \to 0} \frac{\overline{h}}{h}$

Consider making $h$ go to $0$ using only real numbers. Since the complex conjugate of a real number is itself, we have:

${f}^{p} r i m e \left(z\right) = {\lim}_{h \to 0} \frac{\overline{h}}{h} = = {\lim}_{h \to 0} \frac{h}{h} = = {\lim}_{h \to 0} 1 = 1$

Now, make $h$ go to $0$ using only pure imaginary numbers (numbers of the form $a i$). Since the conjugate of a pure imaginary number $w$ is $- w$, we have:

${f}^{p} r i m e \left(z\right) = {\lim}_{h \to 0} \frac{\overline{h}}{h} = = {\lim}_{h \to 0} - \frac{h}{h} = = {\lim}_{h \to 0} - 1 = - 1$

And therefore $f \left(z\right) = \overline{z}$ has no derivative.