Conditions required for (z_(1)z_(2))^(omega)=z_(1)^(omega)z_{2}^(omega), where z_(1),z_(2),omega in bbb(C) I am having trouble finding the conditions on z_1 and z_2 in order for:

Julia Chang 2022-09-27 Answered
Conditions required for ( z 1 z 2 ) ω = z 1 ω z 2 ω , where z 1 , z 2 , ω C
I am having trouble finding the conditions on z 1 and z 2 in order for:
( z 1 z 2 ) ω z 1 ω z 2 ω ω C
My first step was to rewrite the equation as:
e ω Log ( z 1 z 2 ) = e ω Log ( z 1 ) e ω Log ( z 2 )
We have that:
Log ( z 1 z 2 ) Log ( z 1 ) + Log ( z 2 ) + 2 n i π , n Z
And therefore, equating both sides of the equation, we get:
Log ( z 1 ) + Log ( z 2 ) + 2 n i π = Log ( z 1 ) + Log ( z 2 )
However, this appears to me to be true z 1 , z 2 C , yet if we set z 1 = z 2 = 1 and ω = i, then the equality does not hold, so I'm not sure where I'm making my error.
I believe that the condition required is when ( ( z ) R ( z ) = 0 ) ( z ) 0 because this is the point at which Log ( z ) is discontinuous, however, I am unable to prove that this is the case.
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Answers (1)

devilvunga
Answered 2022-09-28 Author has 14 answers
"I am having trouble finding the conditions on z 1 and z 2 in order for:
( z 1 z 2 ) ω z 1 ω z 2 ω ω C "
This question makes no sense as formulated since ω z ω does not define a function on C except for some very specific values of z. Fortunately, the question is soon reformulated as:
"Find some conditions on z 1 and z 2 to ensure that, for every ω in C ,
e ω Log ( z 1 z 2 ) = e ω Log ( z 1 ) e ω Log ( z 2 ) "
and this new version does make sense. Thus, one asks that, for every ω in C
ω Log ( z 1 z 2 ) ω Log ( z 1 ) ω Log ( z 2 )
is an integer multiple of 2 i π. Define a function n on C by
n ( ω ) = ω Log ( z 1 z 2 ) ω Log ( z 1 ) ω Log ( z 2 ) 2 i π .
Then the function n is continuous, integer valued, and n ( 0 ) = 0. This shows that n ( ω ) = 0 for every ω, for example, n ( 1 ) = 0, that is,
Log ( z 1 z 2 ) = Log ( z 1 ) + Log ( z 2 ) .
This necessary condition is obviously sufficient. It holds if and only if
π < Arg ( z 1 ) + Arg ( z 2 ) π .
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