Some questions in Complex Analysis

  • Thread starter kakarotyjn
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  • #1
I'm not very clear of the problems below,so I may make some mistakes,if you point out them and explain to me,I'm reallly grateful.

1.If f(z) is an analytic function,why can we derivate it as a real function to get it's derivation?
I mean f'(z) should be [tex]f^' (z) = \frac{{\partial u}}{{\partial x}} + i\frac{{\partial v}}{{\partial x}}[/tex],we can get the derivation by this formula,but why can we just derivate it as a real function?For example,if [tex]f(z) = \log (z - a)[/tex],then it's derivation is
[tex]f'(z) = \frac{1}{{z - a}}[/tex]?

2.What on earth is principle-valued branch?
Why (z)^(1/2) is multiple-valued?Why we may choose for [tex]\Omega [/tex] the complement of the negative real axis z<=0 then it is a single-valued function?
I'm really confused of it.And why once the continuity is established the analyticity follows by derivation of the inverse function?
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Answers and Replies

  • #2
1. That's the result, it's not a definition. You start with the definition of the derivative and then you get that result. It's not surprising though, since the definition of derivative in complex analysis is the same as in real analysis, except you put dz instead of dx. But the analyticity constraint makes the derivative independent of the direction you take dz to approach zero, therefore you expect the result to be like in real analysis. But to conclude, this is the result of the derivative identity (which a result of the analyticty constraint), try it yourself.

2. The complex power is defined in this way:


Because we know how to deal with exponents and logarithms, but the idea of a complex (or even a real irrational) power is obscure to us.

Since the log itself is a multivalued function (since it contains the arg() function as the imaginary part), the power function will be multivalued itself.

The principle branch of the log(z) function, also denoted Log(z) will also define the principle branch of [tex]f(z)=z^{\alpha}[/tex]

And since all of the problems of Arg() function arise in the negative real axis, these problems will propagate through the definitions and occur in z^a as well.
  • #3
Hi elibj123 ,thanks very much for your explanation.

One more question,why all of the problems of Arg() function arise in the negative real axis? Why not the positive real axis?