# Finding unit vector

1. Mar 12, 2013

### 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

2. Mar 12, 2013

### voko

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

3. Mar 12, 2013

### haruspex

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

4. Mar 12, 2013

### yungman

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 $B^2=B\cdot B^*$?

5. Mar 12, 2013

### yungman

Thank you.

6. Mar 12, 2013

### 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.

7. Mar 12, 2013

### yungman

Thank you very much. I see.