Complex power raised over real number.

AI Thread Summary
Raising a complex power over a real number is a valid operation, as demonstrated in the discussion. The method of expressing a complex power, such as 3 raised to the 5i, using Euler's formula is confirmed to be correct. It is recommended to maintain higher precision in numerical computations to avoid inaccuracies. The conversation also highlights the existence of infinitely many solutions based on the choice of branch or argument in complex analysis. Overall, the topic emphasizes the importance of understanding complex exponentiation and its implications in calculations.
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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