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.
