Complex numbers exponential: Difference between revisions

Revision as of 09:40, 25 December 2006

Revision as of 14:28, 25 December 2006

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{{Importance|date=December 2006}}

{{incoherent}}<!--I can't find a better tag to say "This is in jargon; please rewrite in English prose.-->

Exponentian is a mathematical operation that involving two numbers,the base x and the y:-

:<math>x^y = \underbrace{x \times \cdots \times y}_n</math>

It means we multiply the exponent copies of the base. if we take it from this Concept that well lead us that the exponent must be a counting numbers in other words Natural Numbers {1,2,3,4,...}. So what if we have a negative or a rational or a real or a complex number for exponent , then we have look at it in an other way '''without''' invaliding the basic Concept . Our topic in here is how can we find the expiation of tow complex numbers.

At the beginning we let take X ,Y ,z as complex numbers :-

:<math> x = r(\cos(\theta)+ i \sin(\theta)) = r e^{i\theta} \,</math>

:<math>y = k(\cos(\varphi)+ i \sin(\varphi)) =k e^{i\varphi}= A + B i</math>

:<math>Z = q(\cos(\phi)+ i \sin(\phi)) =q e^{i \phi} \,</math>

where : r,k,q are the absolute values or also called modulus and <math>\theta ,\varphi , \phi</math> are argument values

Then :

Then : x power by y we get :-

:<math> x^{y} = (r e^{i\theta})^{(k e^{i\varphi})} = e^{k[(\cos(\varphi) \ln(r)+ \sin(\varphi) \theta)+ i (\cos(\varphi) \theta - \sin(\varphi)\ln(r))]} \,</math>

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:<math>= {\ln(r)+ i \theta \over\ln(q)+ i \phi} \, </math>

:<math>= {\ln(r e^{i\theta})\over\ln(q e^{i\phi})} = {(\ln(q) \ln(r)+ \theta \phi)+i(\phi \ln(r) - \theta \ln(q) ) \over (\ln(r))^{2}+ (\theta)^{2}}</math>

We going to say that <math>g = (\ln(r))^{2}+ (\theta)^{2} </math> and we continue ...