Continuous complex function without antiderivativeIt's a well-known result that every real continuos function has an antiderivative. Is this theorem still true for a complex function? If not, can someone point out a counter-example (and proof that it is indeed a counter-example)?

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Continuous complex function without antiderivativeIt&#039;s a well-known result that every real continuos function has an antiderivative. Is this theorem still true for a complex function? If not, can someone point out a counter-example (and proof that it is indeed a counter-example)?

ladaroh · Accepted Answer

Step 1The modulus function,   f  (  z  )  =      |    z      |  , is continuous but not complex-differentiable.Suppose it had an antiderivative F, so that       F    ′    (  z  )  =  f  (  z  ). Then F would be infinitely differentiable (because all complex-analytic functions on C are), which would make f differentiable.Why is f not complex-differentiable?Step 2Write   f  (  z  )  =  u  (  z  )  +  i  v  (  z  ); then v is the constant function 0. Then if f were analytic, the Cauchy-Riemann equations would tell you that the derivative of u wrt the real and imaginary parts of z are both 0; thus u is a constant function. But u is not constant, so this is a contradiction.

Continuous complex function without antiderivative It's a well-known result that every

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