As I have been reading about cardinalities a little, I was wondering what is the cardinality of number of values of complex logarithm, which are, as well known, infinite in number with a cycle +- 2pi*I, and made a short training excercise.

If we have a circle with imaginary diameter I, it has a circumference I*pi (2*ln(I))=ln(i^2)

If we have a circle with imaginary diameter I/2, if has circumference I*pi/2 = ln(I)

Circle with imaginary diameter I/4 , circumference I*pi/4= 1/2 ln(I) = ln (sqrt(I))

Circle with imaginary diameter I/8, circumference I*pi/8=1/4*ln(I)= ln(I^(1/4))

Circle with imaginary diameter I/(2^n), circumference I*pi/(2^n)=1/2^(n-1)*ln(I)=ln(I^(1/2^(n-1))

Now if we sum circumferences of infinite number of such progression of circles, we get

Pi*I*( 1+1/2+1/4+1/8+ +1/2^(n-1)+..)= 2*pi*I = 4*ln(I)= ln(I^4) = ln(1) by the sum of infinite series

So we can assume that if we would have added to each I/(2^n) its infinite number of periodic values in form + - (2*pi*k)/n we would have sum of circumferences exactly Ln(1) :

Ln(1) = 0, +-2**I*pi, +-4*I*pi, +- 6 *I*pi ............2*pi*I*n. n natural number

on other hand we have series of logarithms of roots of I which also sums up to the same value:

Ln(-1)+ ln(sqrt(I) + ln (I^(1/4)) + ln (I^(1/) + ln(I^1/16))+ .. =

Ln(-1) + ln(sgrt(I)*(I^(1/4)* (I^(1/)*I^(1/16)* ....)

Obviosly, also right logarithm including roots must sum to Ln(-1). , then Ln(-1)+Ln(-1) = Ln(-1*-1)= Ln(1) as above.

But if we look at the structure of the roots we know that n-th root has n values, so the number of possible combinations of roots in product grows as 2^(n*(n+1)/2)) -triangular numbers, or Pascal triangle.

(sgrt (I) has 2 values each of which can combine with 4 values of 4th root of I which makes totally 8 combinations which can combine with 8 values of 8th root of I so there are 2^6 = 64 combinations which combined with 16 values of 16th roots (2^4) give 2^10 = 1024 combinations which combine with 32 values of 32th root (2^5) gives 2^10*2^5= 2^15 combinations etc.).

So in fact, the powers of 2 of number of combinations are triangular numbers. It could be that some roots in this multiplication cancel out so some paths become equal and the number of effective combinations is reduced?

Now, we can have infinite number of ways we can construct converging or diverging values for complex logarithm as sum of circles with imaginary radiuses , and each of them, when expressed as product of roots of I, will have differerent tree structure behind them.

So what is the cardinality of infinite values of complex logarithm?

Ivars

If we have a circle with imaginary diameter I, it has a circumference I*pi (2*ln(I))=ln(i^2)

If we have a circle with imaginary diameter I/2, if has circumference I*pi/2 = ln(I)

Circle with imaginary diameter I/4 , circumference I*pi/4= 1/2 ln(I) = ln (sqrt(I))

Circle with imaginary diameter I/8, circumference I*pi/8=1/4*ln(I)= ln(I^(1/4))

Circle with imaginary diameter I/(2^n), circumference I*pi/(2^n)=1/2^(n-1)*ln(I)=ln(I^(1/2^(n-1))

Now if we sum circumferences of infinite number of such progression of circles, we get

Pi*I*( 1+1/2+1/4+1/8+ +1/2^(n-1)+..)= 2*pi*I = 4*ln(I)= ln(I^4) = ln(1) by the sum of infinite series

So we can assume that if we would have added to each I/(2^n) its infinite number of periodic values in form + - (2*pi*k)/n we would have sum of circumferences exactly Ln(1) :

Ln(1) = 0, +-2**I*pi, +-4*I*pi, +- 6 *I*pi ............2*pi*I*n. n natural number

on other hand we have series of logarithms of roots of I which also sums up to the same value:

Ln(-1)+ ln(sqrt(I) + ln (I^(1/4)) + ln (I^(1/) + ln(I^1/16))+ .. =

Ln(-1) + ln(sgrt(I)*(I^(1/4)* (I^(1/)*I^(1/16)* ....)

Obviosly, also right logarithm including roots must sum to Ln(-1). , then Ln(-1)+Ln(-1) = Ln(-1*-1)= Ln(1) as above.

But if we look at the structure of the roots we know that n-th root has n values, so the number of possible combinations of roots in product grows as 2^(n*(n+1)/2)) -triangular numbers, or Pascal triangle.

(sgrt (I) has 2 values each of which can combine with 4 values of 4th root of I which makes totally 8 combinations which can combine with 8 values of 8th root of I so there are 2^6 = 64 combinations which combined with 16 values of 16th roots (2^4) give 2^10 = 1024 combinations which combine with 32 values of 32th root (2^5) gives 2^10*2^5= 2^15 combinations etc.).

So in fact, the powers of 2 of number of combinations are triangular numbers. It could be that some roots in this multiplication cancel out so some paths become equal and the number of effective combinations is reduced?

Now, we can have infinite number of ways we can construct converging or diverging values for complex logarithm as sum of circles with imaginary radiuses , and each of them, when expressed as product of roots of I, will have differerent tree structure behind them.

So what is the cardinality of infinite values of complex logarithm?

Ivars