Finding Unit Vector: Calculating|𝑬| & 𝐸̂

[yungman]

I want to find the unit vector of $\vec E= \hat x A\;+\;\hat y Be^{j\phi}$

$\hat E=\frac {\vec E}{|\vec E|}$

From my work: $|\vec E|=\sqrt{A^2+(Be^{j\phi})^2}$

My question is what is $(Be^{j\phi})^2$?

Do I substitude $e^{j\phi}=\cos \phi +j\sin \phi$? So $(Be^{j\phi})^2=B^2[(\cos\phi+j\sin\phi)(\cos\phi-j\sin\phi)]\;=\;B^2(\cos^2\phi+\sin^2\phi)\;=\;B^2$

$\Rightarrow\;|\vec E|=\sqrt{A^2+(Be^{j\phi})^2}\;=\;\sqrt{A^2+B^2}$ and

$$\hat E\;=\;\frac{\hat x A\;+\;\hat y Be^{j\phi}}{\sqrt{A^2+B^2}}$$

Thanks

 

[voko]

That is correct. Any value of the imaginary exponential is on the unit circle in the complex plane, so its magnitude is unity.

 

[haruspex]

$(Be^{j\phi})^2$?

$(Be^{j\phi})^2 = B^2e^{2j\phi}$, but what you want is $|Be^{j\phi}|^2 = B^2$

 

[yungman]

$(Be^{j\phi})^2 = B^2e^{2j\phi}$, but what you want is $|Be^{j\phi}|^2 = B^2$

Thanks for the reply. So what you mean is:

$|\vec E|=\sqrt{A^2+|Be^{j\phi}|^2}$

This is the main confusion for me. If you look at the ordinary way of finding the magnitude of a vector with real value, $|\vec E|=\sqrt{(A)^2+(B)^2}$. So for complex vector we use "absolute" value...which means $B^2=B\cdot B^*$?

 

[yungman]

That is correct. Any value of the imaginary exponential is on the unit circle in the complex plane, so its magnitude is unity.

Thank you.

 

[voko]

The magnitude of complex vectors, just like that of real vectors, is defined via their inner (dot, scalar) product. Recall how the inner product of complex vectors is defined.

 

[yungman]

The magnitude of complex vectors, just like that of real vectors, is defined via their inner (dot, scalar) product. Recall how the inner product of complex vectors is defined.

Thank you very much. I see.