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(z)#, 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^prime(z) = lim_(h to 0) (f(z+h)-f(z))/(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(z) = bar(z)#, that is, #f# takes a complex number #z# into it's conjugate #bar(z)#.

Then, the derivative of #f# is

#f^prime(z) = lim_(h to 0) (f(z+h)-f(z))/(h) = lim_(h to 0) (bar(z+h)-bar(z))/(h) = lim_(h to 0) (bar(h) + bar(z)-bar(z))/(h) = lim_(h to 0) (bar(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^prime(z) = lim_(h to 0) (bar(h))/(h) = = lim_(h to 0) h/h = = lim_(h to 0) 1 = 1#

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

#f^prime(z) = lim_(h to 0) (bar(h))/(h) = = lim_(h to 0) -h/h = = lim_(h to 0) -1 = -1#

And therefore #f(z) = bar(z)# has no derivative.