You can treat ii as any constant like CC. So the derivative of ii would be 00.
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)f(z), where zz is a complex number (that is, ff has a complex domain). Then the derivative of ff 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.
