simplification of an complex exponential equation

Question

There are these steps in a solutions manual I do not follow. I struggle to find any good and problem specific information about this kind of math wizardry on my own. I don't really know what to google for... How did they get rid of the complex exponential? And how did it return again?

I'm sorry, not very good with words either ;)

And I'm unable to determine which tag is most appropriate...

Answer 1

We can write this in component form by using Euler's Formula: $$e^{j\theta}=\cos\theta+j\sin(\theta)$$

Where $j$ is the imaginary unit as written in electrical engineering, i.e. $j^{2}=-1$. If we then take the principle square root to be $j=i$, we can write your expression as:

$$\frac{1+j}{1-\frac{1}{2}}=2(1+j)$$

We can then convert this to polar form using the facts that $|2(1+j)|=\sqrt{2^{2}+2^{2}}=\sqrt{8}=2\sqrt{2}$ and $\varphi=\tan^{-1}(1)=\frac{\pi}{4}$, which gives us:

$$\frac{1-e^{-j\frac{\pi}{2}}}{1+\frac{e^{-j\pi}}{2}}=2\sqrt{2}e^{j\frac{\pi}{4}}$$

Answer 2

Assuming $\;j=\sqrt{-1}=:i\;$ :

$$\frac{1-e^{-i\pi/2}}{1+\frac12e^{-i\pi}}=\frac{1-(-i)}{1+\frac12(-1)}=2(1+i)=2\sqrt2e^{i\pi/4}$$

since

$$|2(i+i)|=\sqrt{4+4}=2\sqrt2\;\;,\;\;\arctan\frac22=\arctan1=\frac\pi4$$

the last one following from the fact that both the real and imaginary parts are positive.

Answer 3

First thing to realise is that here, $j$ is being used to denote the complex number that we often denote by $i$; note that the use of $j$ is common in electrical engineering.

By using Euler's identity $e^{j\theta} = \cos\theta + j\sin\theta$, or thinking geometrically, you can determine $e^{-j\frac{\pi}{2}}$ and $e^{-j\pi}$. Substituting these values in, the expression should reduce to $2(1+j)$, not $2(1-j)$.

Note that $2(1+j)$ is the Cartesian expression for $H(e^{j\frac{\pi}{4}})$, but we could also write it in polar form. Note that $|2(1+j)| = 2\sqrt{1^2+1^2} = 2\sqrt{2}$, so $2(1+j) = 2\sqrt{2}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}j\right)$. By using Euler's identity, or thinking geometrically, we have $\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}j = e^{j\frac{\pi}{4}}$ so $2(1+j) = 2\sqrt{2}e^{j\frac{\pi}{4}}$.