Find an antiderivative for z I'm having some trouble on this homework problem:If z is...

Leonel Contreras

Leonel Contreras

Answered

2022-07-01

Find an antiderivative for z
I'm having some trouble on this homework problem:
If z is defined by z = e ( log z ) / 2 for the branch of the log function defined by the condition π / 2 arg ( z ) 3 π / 2, find an antiderivative for z and then find γ z d z, where γ is any path from -1 to 1 which lies in the upper half-plane.
I think the second part will be relatively straightforward by FTC once I find an antiderivative. What I'm having trouble with is the first part. My intuition from real-valued calculus says that an antiderivative for z is 2 / 3 z 3 / 2 . I'm not sure how to show this in the complex setting though. How do I use the log function and the associated branch to show this? Thanks.

Answer & Explanation

Rebekah Zimmerman

Rebekah Zimmerman

Expert

2022-07-02Added 32 answers

Step 1
We let our intuition tell us that we expect the antiderivative is g ( z ) = 2 3 z 3 2
We can write this in polar form and expand it to
g ( z ) = 2 3 e 3 2 log ( r ) cos ( 3 2 θ ) + i ( 2 3 e 3 2 log ( r ) sin ( 3 2 θ ) ) = u ( r , θ ) + i v ( r , θ )
where r = | z | and θ = arg ( z ).
Then, taking partial derivatives yields
u r = r 1 2 cos ( 3 2 θ ) ; u θ = r 3 2 sin ( 3 2 θ ) ; v r = r 1 2 sin ( 3 2 θ ) ; v θ = r 3 2 cos ( 3 2 θ )
We can then easily verify that the Cauchy-Riemann equations hold, namely (for polar coordinates) u r = v θ r and u θ = r v r .
Step 2
Since CR holds, we know that the function g(z) is analytic and has derivative given by g ( z ) = [ cos ( θ ) u r sin ( θ ) u θ r ] + i [ cos ( θ ) v r sin ( θ ) v θ r ]
Plugging in and following the algebra, we are left with g ( z ) = r 1 2 e i θ 2 = z . Thus, we've verified that g(z) is indeed an antiderivative for z .
Next, we let γ : [ 1 , 1 ] C be a path in the upper half-plane.
By the Fundamental Theorem of Calculus, we know γ f ( z ) d z = g ( γ ( b ) ) g ( γ ( a ) ), where g is an antiderivative for f.
Applying this here, we get γ z d z = 2 3 ( 1 ) 3 2 2 3 ( 1 ) 3 2 = 2 3 2 3 ( i 3 ) = 2 3 + 2 3 i
Kendrick Hampton

Kendrick Hampton

Expert

2022-07-03Added 7 answers

Step 1
Your problem would need to read π 2 < a r g ( z ) < 3 π 2 since including the possibility that one of the signs can be equal would mean z 1 2 is not continuous (and thus not holomorphic on the domain of definition), and including both possibilities for equality would mean that z 1 2 is not well-defined.
Step 2
Your intuition is correct. To prove this, use the definition of derivative with respect to z for complex functions on F ( z ) = 2 3 z 3 2 . This will show that F(z) is an antiderivative of z 1 2 on the domain of definition. The procedure works out identically to how the limit calculation would work in a calculus class.
To finish, note that F(z) is itself holomorphic on the given domain of definition (why?). In such cases, it is true that for any C 1 path γ ( t ) C over [a,b] contained in the (open, connected) domain of definition for a holomorphic function F(z), we have F ( γ ( B ) ) F ( γ ( A ) ) = a b F ( γ ( t ) ) γ ( t ) d t = γ F ( z ) d z .

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