Logarithm rules for complex numbersAre the logarithm rules true for complex numbers?We know that for positive real numbers a, b, c and real number d that: log b ⁡ ( a d ) = d log b ⁡ ( a ) log b ⁡ ( a ) = log c ⁡ ( a ) log c ⁡ ( b ) log b ⁡ ( x y ) = log b ⁡ ( x ) + log b ⁡ ( y ) log b ⁡ ( x y ) = log b ⁡ ( x ) − log b ⁡ ( y )We also know that log b ⁡ ( b d ) = dDoes this extend to complex numbers as a, b, c or d?My instinct is that log b ⁡ ( b s + t i ) = s + t i In other words, I'm pretty confident that the last formula works when d is a complex number

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Logarithm rules for complex numbersAre the logarithm rules true for complex numbers?We know that for positive real numbers   a,   b,   c and real number   d that:      log    b    ⁡      (          a      d        )    =  d      log    b    ⁡  (  a  )      log    b    ⁡  (  a  )  =                    log        c            ⁡      (      a      )                      log        c            ⁡      (      b      )            log    b    ⁡  (  x  y  )  =      log    b    ⁡  (  x  )  +      log    b    ⁡  (  y  )      log    b    ⁡      (          x      y        )    =      log    b    ⁡  (  x  )  −      log    b    ⁡  (  y  )We also know that      log    b    ⁡      (          b      d        )    =  dDoes this extend to complex numbers as   a,   b,   c or   d?My instinct is that      log    b    ⁡      (          b              s        +        t                  i                      )    =  s  +  t      i  In other words, I&#039;m pretty confident that the last formula works when   d is a complex number

Anika Stevenson · Accepted Answer

You have to be careful because logs and non-integer powers are multivalued functions. The definition is that       a    d    =  exp  ⁡  (  d  ln  ⁡  (  a  )  ) (for any branch of   ln). Now       log    b    ⁡  (      a    d    ) is any   z such that       b    z    =      a    d  , i.e.   exp  ⁡  (  z  ln  ⁡  (  b  )  )  =  exp  ⁡  (  d  ln  ⁡  (  a  )  ), and that is equivalent to   z  ln  ⁡  (  b  )  −  d  ln  ⁡  (  a  )  =  2  π  i  n for some integer   n. So the result is      log    b    ⁡  (      a    d    )  =                    d        ln        ⁡        (        a        )        +        2        π        i        n                    ln        ⁡        (        b        )            Similarly,      log    b    ⁡  (  a  )  =                    ln        ⁡        (        a        )        +        2        π        i        m                    ln        ⁡        (        b        )            for some integer   m. And thus (assuming you use the same values of   ln  ⁡  (  a  ) and   ln  ⁡  (  b  ) in both cases)      log    b    ⁡  (      a    d    )  −  d      log    b    ⁡  (  a  )  =  2  π  i                    n        −        m        d                    ln        ⁡        (        b        )            For example, take   a  =  b  =  e and   d  =  2  π  i, and use the principal branch of   ln      log    e    ⁡  (      e          2      π      i        )  =  ln  ⁡  (  1  )  =  0 but   2  π  i      log    e    ⁡  (  e  )  =  2  π  iAnother interesting example is   a  =  b  =  −  1,   d  =  3. Now   (  −  1      )    3    =  −  1, but there is no way to have       log          −      1        ⁡  (  −  1  )  =      log          −      1        ⁡  (  (  −  1      )    3    )  =  3      log          −      1        ⁡  (  −  1  ) (this would imply       log          −      1        ⁡  (  −  1  )  =  0, which is certainly false).

Logarithm rules for complex numbers Are the logarithm rules true for complex numbers? We know that

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