# Euler Representation of complex numbers

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1. Jan 8, 2016

### Hijaz Aslam

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?

Last edited: Jan 8, 2016
2. Jan 8, 2016

3. Jan 8, 2016

### Hijaz Aslam

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?

4. Jan 8, 2016

### suremarc

Indeed, your mistake is on the first line. Remember that $Re(z)=|z|\cos\theta$. Then we have
$$|r_1e^{i\theta_1}+r_2e^{i\theta_2}|^2=|r_1e^{i\theta_1}|^2+|r_2e^{i\theta_2}|^2+2r_1r_2\cos(\theta_1-\theta_2)$$

5. Jan 8, 2016

### Hijaz Aslam

Now I want to know how that angle $\theta_1-\theta_2$ crept into the equation? Can you elaborate?

6. Jan 8, 2016

### suremarc

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?

7. Jan 9, 2016

### Hijaz Aslam

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.