Euler Representation of complex numbers

Hijaz Aslam
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I am bit confused with the Eueler representation of Complex Numbers.

For instance, we say that e^{i\pi}=cos(\pi)+isin(\pi)=-1+i0=-1.
The derivation of e^{i\theta}=cos(\theta)+isin(\theta) is carried out using the Taylor series. I quite understand how ##e^{i\pi}## turns out to be ##-1## using taylor series. But what is the mathematical meaning of ##e^{i\pi}##? How can a constant (##e##) be raised to an 'entity' like ##i=\sqrt{-1}##?

This problem started to concern me when I tried the following out.
A theorem states that : |z_1+z_2|^2=|z_1|^2+|z_2|^2+2Re(z_1\bar{z_2})=|z_1|^2+|z_2|^2+2|z_1||z_2|cos(\theta_1-\theta_2)

But I tried solving this out using the Euler number like: |z_1+z_2|^2=|(z_1+z_2)^2|=|(r_1e^{i\theta_1}+r_2e^{i\theta_2})^2|= |r_1e^{i\theta_1}|^2+|r_2e^{i\theta_2}|^2+|2r_1r_2e^{i(\theta_1+\theta_2)}|=r_1+r_2+2r_1r_2=|z_1|^2+|z_2|^2+2|z_1||z_2|

I know that am seriously wrong somewhere. Can I follow out the "complex" algebra of 'complex numbers' by using Euler's form in simple algebra?
 
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Alright, I think I've made a 'grand' mistake by stating: |(r_1e^{i\theta_1}+r_2e^{i\theta_2})^2|= |r_1e^{i\theta_1}|^2+|r_2e^{i\theta_2}|^2+|2r_1r_2e^{i(\theta_1+\theta_2)}|.

Of course |(r_1e^{i\theta_1}+r_2e^{i\theta_2})^2|= (r_1e^{i\theta_1})^2+(r_2e^{i\theta_2})^2+2r_1r_2e^{i(\theta_1+\theta_2)}|\neq|r_1e^{i\theta_1}|^2+|r_2e^{i\theta_2}|^2+|2r_1r_2e^{i(\theta_1+\theta_2)}|.

So, how do I get along using the Euler Form?
 
Hijaz Aslam said:
Alright, I think I've made a 'grand' mistake by stating: |(r_1e^{i\theta_1}+r_2e^{i\theta_2})^2|= |r_1e^{i\theta_1}|^2+|r_2e^{i\theta_2}|^2+|2r_1r_2e^{i(\theta_1+\theta_2)}|.
Indeed, your mistake is on the first line. Remember that ##Re(z)=|z|\cos\theta##. Then we have
|r_1e^{i\theta_1}+r_2e^{i\theta_2}|^2=|r_1e^{i\theta_1}|^2+|r_2e^{i\theta_2}|^2+2r_1r_2\cos(\theta_1-\theta_2)
 
suremarc said:
Indeed, your mistake is on the first line. Remember that ##Re(z)=|z|\cos\theta##. Then we have
|r_1e^{i\theta_1}+r_2e^{i\theta_2}|^2=|r_1e^{i\theta_1}|^2+|r_2e^{i\theta_2}|^2+2r_1r_2\cos(\theta_1-\theta_2)

Now I want to know how that angle ##\theta_1-\theta_2## crept into the equation? Can you elaborate?
 
Hijaz Aslam said:
Now I want to know how that angle ##\theta_1-\theta_2## crept into the equation? Can you elaborate?
Sorry, I was hasty in answering. I'll backtrack a bit.

Your mistake is in misapplying the absolute value. Recall that ##|z|=\sqrt{z\bar{z}}##, so that ##|z|^2=z\bar{z}##. This turns ##|z+w|^2## into the product ##(z+w)(\bar{z}+\bar{w})##, which can be expanded by distributivity.

As for your earlier question--imagine that ##e^{it}## is the position of a point mass, where the real and imaginary axes replace the x- and y-axes, respectively. Then compare the tangent vector with the complex derivative. What do you see?
 
Oh yes! Thanks a lot suremac. I've missed out that point. So, I presume there isn't much to do by taking
|(r_1e^{i\theta_1}+r_2e^{i\theta_2})^2|=|(r_1e^{i\theta_1})^2+(r_2e^{i\theta_2})^2+2r_1r_2e^{i(\theta_1+\theta_2})| rather than getting confused.

Am afraid that I don't understand the question you have posed. We represent a complex number in a complex plane by a vector whose magnitude is ##|z|##. I understand that ##e^{i\theta}## a sort of function defined by : f(x)=e^{i\theta}=cos\theta+isin\theta and by plugging in the value ##\theta=\pi## yields an outcome of ##-1##, i.e ##f(\pi)=-1##. Just like we represent any other function.
But I am little confused with the "non-functional" value of ##e^{i\theta}## that is, the numerical value of ##2.17^{i\pi}## (like ##2^3=8##) etc. Am I confusing a property intrinsic to real numbers alone with a 'complex attribute'? I think I am indirectly questioning the 'concievable' numerical value of ##i##. Sorry if I am being irrational.
 

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