# How to Convert a Complex Logarithm to a Complex Exponential

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• CalcExplorer
In summary: If the above is ##z^k+iC##, there's no way that taking the log would be helpful, as there is no property that can break up ##\log(A + B)##.Or did you mean to write ##z^{k + iC}##?In that case ##\log(z^{k + iC}) = (k + iC)\log(z)##.
CalcExplorer
Okay, so I'm working with a rather frustrating problem with a calculus equation. I'm trying to solve a calculus equation which I conceptualized from existing methods involving complex number fractal equations. I'm very familiar with pre-calculus, while being self-taught in portions of calculus for practical applications in coding and higher dimensional mathematics, so bare with me on this.

Here's the premise:
I'm using the Mandelbrot Equation [ z = z2 + i C ] and the Hausdorff Dimension [ N = sd ], where d = ln(N)/ln(s), to create a new iterational equation, which fractally conforms to an already defined Hausdorff Dimension [ log(20)/log(2+φ) ], where φ = ((√5)+1)/2. This specific Hausdorff Dimension is of a Dodecahedron Fractal Flake.

This provides the new equation:
z = zlog(20)/log(2+φ) + i C

I'm attempting to solve the first iteration of this new equation in quadratic form, the format the mandelbrot is solved in [ z2 = a2-b2 + 2abi ], as it's the format most advantageous for the graphical mapping of the fractal structure.

From what I've surmised thus far in order to solve such an equation I need to first convert the complex logarithm into a complex exponential using Euler's Formula, and then solve the new formula algebraically to derive the first iterational solution.

There seems to be a basis for this method, albeit with certain conditions in the solutions, and similar questions put up on this forum before, but I don't quite understand the principles enough to solve it myself.

These are the relevant mathematical references on the topic I've been able to find -
http://math.gmu.edu/~rsachs/m114/eulerformula.pdf

But that's pretty much it and I don't quite grasp how it's converted, since there's no direct examples of this process that I could find with conversions using Euler's Formula and Complex Logarithms.

Is this the proper way to go about solving such a problem?
And is there someone here that can help to find the solution?

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I'm not sure I understand what it is that you're doing, or what it is that you're asking. The number ##log(20)/log(2+\varphi)## is just a number, which you can calculate. Let's call it ##k##. Are you just asking how to compute ##z^k + iC##?

I don't know immediately what the relationship is between the fractal dimension of the iteration and the power ##k##. I don't think the two are necessarily connected.

stevendaryl said:
I'm not sure I understand what it is that you're doing, or what it is that you're asking. The number ##log(20)/log(2+\varphi)## is just a number, which you can calculate. Let's call it ##k##. Are you just asking how to compute ##z^k + iC##?
Yes, I'm aware that it's a calculable number, k = 2.3296217161703454689697018540751, but that's not the ideal format to compute ##z^k + iC##. As I'm trying to define it's first solvable iteration in quadratic form.
stevendaryl said:
I don't know immediately what the relationship is between the fractal dimension of the iteration and the power ##k##. I don't think the two are necessarily connected.
This is a problem which was published before in a very interesting thesis, which also describes it's relationship between the fractal dimension of the iterations and the power ##k## :
https://www.math.hmc.edu/seniorthesis/archives/2003/shaas/shaas-2003-thesis.pdf

CalcExplorer said:
Yes, I'm aware that it's a calculable number, k = 2.3296217161703454689697018540751, but that's not the ideal format to compute ##z^k + iC##. As I'm trying to define it's first solvable iteration in quadratic form.
If the above is ##z^k + iC##, there's no way that taking the log would be helpful, as there is no property that can break up ##\log(A + B)##.
Or did you mean to write ##z^{k + iC}##?
In that case ##\log(z^{k + iC}) = (k + iC)\log(z)##.
CalcExplorer said:
This is a problem which was published before in a very interesting thesis, which also describes it's relationship between the fractal dimension of the iterations and the power ##k## :
https://www.math.hmc.edu/seniorthesis/archives/2003/shaas/shaas-2003-thesis.pdf

Mark44 said:
If the above is ##z^k + iC##, there's no way that taking the log would be helpful, as there is no property that can break up ##\log(A + B)##.
Or did you mean to write ##z^{k + iC}##?
In that case ##\log(z^{k + iC}) = (k + iC)\log(z)##.

Sorry, the confusion is from an error in the OP.

Given the aformentioned thesis the Hausdorff dimension can be expressed in polynomial form.
So, the new equation becomes:

##z=z^k+iC##

where, ##k=z^k+iC## [solved numerically giving the complex number julia set]
k = z(log(20)/log(2+φ))## + 2.329621i##

This means k is a complex logarithm able to be converted into a complex exponential using Eulers formula. Once converted it could be solved but for the z iteration.

At least in theory anyway, using this method would the conversion from a complex logarithm to a complex exponential be valid and possible? If it can be converted, what would be the solution for the quadratic form solution of ##z=z^k+iC##?

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Following the iteration therefore for k:
##k = k##0(log(20)/log(2+φ))##+2.329621i##
##k##0## = 0 →k##1## = 0^k + c ≡c##
##c = 2.329621i##
##k##2## = c##(log(20)/log(2+φ))##+2.329621i##

Which means:
##k = c##(log(20)/log(2+φ))##+2.329621i##

After the conversion of the complex log into a complex exponential the new form of the equation
##k=c##(log(20)/log(2+φ))
is solved in quadratic form, and then used in the ##z=z^k+iC## also solved in quadratic form

Last edited:

## 1. How do you convert a complex logarithm to a complex exponential?

The process of converting a complex logarithm to a complex exponential involves using the properties of logarithms and exponents to rewrite the expression in a different form.

## 2. What are the properties of logarithms and exponents used in this conversion?

The main properties used are the logarithm power rule, which states that logb(xy) = y * logb(x), and the exponential power rule, which states that (bx)y = bxy.

## 3. Can you provide an example of converting a complex logarithm to a complex exponential?

Sure, let's say we have log2(3+4i). Using the logarithm power rule, we can rewrite this as (3+4i) * log2(2). Then, using the exponential power rule, we can rewrite this as 2(3+4i).

## 4. Are there any special cases to consider when converting a complex logarithm to a complex exponential?

Yes, there are a few special cases to be aware of. For example, when the base of the logarithm is the same as the base of the exponential, the conversion is simply a matter of rearranging the expression. Also, when the argument of the logarithm is a negative real number, you may need to use the complex number i in the final exponential expression.

## 5. Why would someone want to convert a complex logarithm to a complex exponential?

Converting a complex logarithm to a complex exponential can be useful in simplifying complex expressions and solving certain types of equations. It can also help in understanding the relationships between logarithms and exponents in a more visual way.

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