What is the derivative of #i#?

Question

Ezra Adams · Accepted Answer

You can treat #i# as any constant like #C#. So the derivative of #i# would be #0#.  However, we have to exercise caution when discussing functions, derivatives, and integrals when working with complex numbers. 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.

William Abdalla · Answer

undefined The derivative of the imaginary unit $ i $ with respect to a real variable is zero, because $ i $ is a constant in the context of real differentiation.