Complex power raised over real number.

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SUMMARY

The discussion centers on the validity of raising a complex power over a real number, specifically the expression \(3^{5i}\). Participants confirm that the operation is permissible and provide a method for expressing the result using Euler's formula: \(3^{5i} = e^{i 5 \ln 3} = \cos(5 \ln 3) + i \sin(5 \ln 3)\). It is emphasized that for numerical computations, maintaining higher precision is crucial, as demonstrated by the approximation \(0.7037573 - 0.7104404i\). The conversation also touches on the existence of infinitely many solutions due to the periodic nature of trigonometric functions.

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PrashntS
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1. I actually don't know if such kind of operation is even allowed.

A friend of mine raised this question, that can we raise a complex power over a real number. I solved it this way. Is this correct?

http://i45.tinypic.com/254vwux.jpg

Homework Statement


Homework Equations


The Attempt at a Solution

 
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Yes, that's fine.
 
Yes, you did it correctly. I am not quite sure approximating the transcandental functions with decimals, but I suppose that's fine.
 
PrashntS said:
1. I actually don't know if such kind of operation is even allowed.

A friend of mine raised this question, that can we raise a complex power over a real number. I solved it this way. Is this correct?

http://i45.tinypic.com/254vwux.jpg

Your method is OK, but if you plan to use the results in further numerical computations, you should keep more digits of accuracy; nowadays, in this computer age, there is no barrier to retaining more digits. For example, it might be better (depending on future uses) to write
3^{5i} = e^{i 5 \ln 3 } = \cos(5 \ln 3) + i \sin(5 \ln 3) \doteq <br /> 0.7037573 - 0.7104404 i\, .

RGV
 
Ray Vickson said:
Your method is OK, but if you plan to use the results in further numerical computations, you should keep more digits of accuracy; nowadays, in this computer age, there is no barrier to retaining more digits. For example, it might be better (depending on future uses) to write
3^{5i} = e^{i 5 \ln 3 } = \cos(5 \ln 3) + i \sin(5 \ln 3) \doteq <br /> 0.7037573 - 0.7104404 i\, .

RGV

Yeah this is obviously much better, I was just looking whether it is even possible or not as none of the books I use has such question.
Would be awesome if you could suggest some good book for complex number (pre collage).
 
Are you also taking into account that there are infinitely-many solutions depending

on your choice of branch/argument?
 
Bacle2 said:
Are you also taking into account that there are infinitely-many solutions depending

on your choice of branch/argument?

well that's obvious, isn't it? trigo fn is periodic, i just stayed in principle branch
 

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