Does the logarithm inequality extend to the complex plane?For estimates, the inequality log ⁡ ( y ) ≤ y − 1 , y &gt; 0 is often helpful. Is there any sort of upper bound for the logarithm function in the complex plane? Specifically, | log ⁡ ( z ) | ≤ something for all z ∈ C Perhaps this would work?: log ⁡ ( z ) ≤ log 2 ⁡ | z | + arg ⁡ ( z ) 2

Question

Does the logarithm inequality extend to the complex plane?For estimates, the inequality   log  ⁡  (  y  )  ≤  y  −  1  ,  y  &amp;gt;  0 is often helpful. Is there any sort of upper bound for the logarithm function in the complex plane? Specifically,       |    log  ⁡  (  z  )      |    ≤ something for all   z  ∈      C  Perhaps this would work?:   log  ⁡  (  z  )  ≤            log      2        ⁡          |        z          |        +    arg    ⁡    (    z          )      2

Genesis Rosario · Accepted Answer

For the principal branch   log  ⁡  z  =  log  ⁡      |    z      |    +  i  arg  ⁡  z   you get the somewhat trivial inequality      |    log  ⁡  z      |    =      |    log  ⁡      |    z      |    +  i  arg  ⁡  z      |    ≤      |    log  ⁡      |    z      |        |    +  πAnother restricted one      |    log  ⁡  (  1  +  z  )      |    ≤  −  log  ⁡  (  1  −      |    z      |    )  ,        |    z      |    &amp;lt;  1

Melina Barber · Accepted Answer

GLOBAL BOUND:If one is seeking a global bound, then we have on the principal branch   log  ⁡  (  z  )  =  Log  (  z  )  =  log  ⁡  (      |    z      |    )  +  Arg  (  z  ), where   −  π  &amp;lt;  Arg  (  z  )  ≤  π, and for       |    z      |    &amp;gt;  0                              |                Log        (        z        )                  |                                    =                  |                log        ⁡        (                  |                z                  |                )        +        i        Arg        (        z        )                  |                                                                ≤                              log            2                    ⁡          (                      |                    z                      |                    )          +                      π            2                                            For       |    z      |    ≥  1, we have                              |                Log        (        z        )                  |                ≤                  (                      |                    z          −          1                      |                                )            2                    +                      π            2                                            while for   0  &amp;lt;      |    z      |    &amp;lt;  1, we have                              |                Log        (        z        )                  |                ≤                                            |                                                z                  −                  1                                z                            |                        2                    +                      π            2                                            A USEFUL LOCAL BOUND:If we restrict       |    z      |   such that       |    z  −  1      |    ≤  ρ for   ρ  ∈  (  0  ,  1  ), then we can find a useful upper bound. We write   Log  (  z  )  =      ∫    0    1              z      −      1              1      +      (      z      −      1      )      t          d  t such that                              |                Log        (        z        )                  |                                    =                  |                      ∫            0            1                                              z              −              1                                      1              +              (              z              −              1              )              t                                          d          t          |                                                                ≤                  ∫          0          1                                                    |                        z            −            1                          |                                                          |                        1            +            (            z            −            1            )            t                          |                                              d        t                                                        ≤                  ∫          0          1                                                    |                        z            −            1                          |                                                          |                        1            −                          |                        z            −            1                          |                        t                          |                                              d        t                                          (1)                                  ≤                                            |                        z            −            1                          |                                            1            −            ρ                              For example, if we take   ρ  =  1      /    2, then       |    Log  (  z  )      |    ≤  2      |    z  −  1      |   for       |    z  −  1      |    &amp;lt;  1      /    2APPLICATION:One application of the inequality   (  1  ) is to prove that the infinite product representation for the sinc function, given by             sin      ⁡      (      π      z      )              π      z        =      ∏          n      =      1        ∞        (    1    −                  z        2                    n        2              )   converges uniformly on compact sets.Equipped with   (  1  ), we have for       |    z      |    &amp;lt;  B and   N  &amp;gt;  B      /        ρ                                |          log          ⁡                      (                          ∏                              n                =                N                            ∞                                      (              1              −                                                z                  2                                                  n                  2                                            )                        )                    |                                    =                  |                      ∑                          n              =              N                        ∞                    log          ⁡                      (            1            −                                          z                2                                            n                2                                      )                    |                                                                ≤                  1                      1            −            ρ                                    ∑                      n            =            N                    ∞                                                    |                                      z              2                                      |                                            n            2                                                                          ≤                              B            2                                1            −            ρ                                    ∑                      n            =            N                    ∞                          1                      n            2                              For any   ϵ  &amp;gt;  0, there exists a number       N    ′   such that whenever   N  &amp;gt;      N    ′  ,       ∑          n      =      N        ∞        1          n      2        &amp;lt;  ϵ  (  1  −  ρ  )      /        B    2  . And hence, the convergence of the infinite product representation of the sinc function is uniform.

Answered question

Answer & Explanation

New Questions in Algebra II