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.

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Find an antiderivative for       z  I&#039;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&#039;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&#039;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.

Rebekah Zimmerman · Accepted Answer

Step 1We 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 2Since 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&#039;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 · Accepted Answer

Step 1Your problem would need to read   −      π    2    &amp;lt;  a  r  g  (  z  )  &amp;lt;            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 2Your 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  .

Find an antiderivative for <msqrt> z </msqrt> I'm having some trouble on this homewor

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