Does the logarithm inequality extend to the complex plane? For estimates, the inequality log(y)<=y−1, y>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 in bbbC Perhaps this would work?: log(z)<=\sqrt(\log^2|z|+arg(z)^2)

Chelsea Lamb 2022-09-27 Answered
Does the logarithm inequality extend to the complex plane?
For estimates, the inequality log ( y ) y 1 , y > 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
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Answers (2)

Genesis Rosario
Answered 2022-09-28 Author has 11 answers
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 | | + π
Another restricted one
| log ( 1 + z ) | log ( 1 | z | ) , | z | < 1
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Melina Barber
Answered 2022-09-29 Author has 3 answers
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 < Arg ( z ) π, and for | z | > 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 < | z | < 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 | < 1 / 2
APPLICATION:
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 | < B and N > 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 > 0, there exists a number N such that whenever N > N , n = N 1 n 2 < ϵ ( 1 ρ ) / B 2 . And hence, the convergence of the infinite product representation of the sinc function is uniform.
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