Simplification of complex expression

[roam]

Homework Statement

For the expression:

$$E=\frac{E_{0}}{2}\left(\exp\left[\frac{j\pi V}{2V_{\pi}}\right]+j\exp\left[-\frac{j\pi V}{2V_{\pi}}\right]\right)$$

I want to show that if $V=m(t)-\frac{V_{\pi}}{2}$, then $|E|^2$ can be written as:

$$|E|^2=\frac{E^2_{0}}{2}\left(1-\cos\left(\frac{\pi m(t)}{V_{\pi}}\right)\right). \tag{1}$$

Note: here $j^2=-1$.

Homework Equations

The Attempt at a Solution

Substituting:

$$E(t)=\frac{E_{0}}{2}\left(\exp\left[\frac{j\pi}{2V_{\pi}}\left(m(t)-\frac{V_{\pi}}{2}\right)\right]+j\exp\left[-\frac{j\pi}{2V_{\pi}}\left(m(t)-\frac{V_{\pi}}{2}\right)\right]\right)$$

$$=\frac{E_{0}}{2}\left(\exp\left[j\left(\frac{\pi}{2V_{\pi}}m(t)-\frac{\pi}{4}\right)\right]+j\exp\left[-j\left(\frac{\pi}{2V_{\pi}}m(t)-\frac{\pi}{4}\right)\right]\right)$$

Multiplying by the complex conjugate:

$|E(2)|^{2}=\left(\frac{E_{0}}{2}\right)^{2}\left(\exp\left[j\left(\frac{\pi}{2V_{\pi}}m(t)-\frac{\pi}{4}\right)\right]+j\exp\left[-j\left(\frac{\pi}{2V_{\pi}}m(t)-\frac{\pi}{4}\right)\right]\right).\left(\exp\left[j\left(\frac{\pi}{2V_{\pi}}m(t)-\frac{\pi}{4}\right)\right]-j\exp\left[-j\left(\frac{\pi}{2V_{\pi}}m(t)-\frac{\pi}{4}\right)\right]\right)$

$$|E(2)|^{2}=\underline{\left(\frac{E_{0}}{2}\right)^{2}\left(\exp\left[j\left(\frac{\pi m(t)}{V_{\pi}}-\frac{\pi}{2}\right)\right]+\exp\left[-j\left(\frac{\pi m(t)}{V_{\pi}}-\frac{\pi}{2}\right)\right]\right)}.$$

Writing this explicitly in terms of trigonometric functions:

$=\left(\frac{E_{0}}{2}\right)^{2}\left[\left(\cos\left(\frac{\pi m(t)}{V_{\pi}}\right)+j\sin\left(\frac{\pi m(t)}{V_{\pi}}\right)\right)\left(\underbrace{\cos\left(-\frac{\pi}{2}\right)+j\sin\left(-\frac{\pi}{2}\right)}_{-j}\right)+\left(\cos\left(-\frac{\pi m(t)}{V_{\pi}}\right)+j\sin\left(-\frac{\pi m(t)}{V_{\pi}}\right)\right)\underbrace{\left(\cos\left(\frac{\pi}{2}\right)+j\sin\left(\frac{\pi}{2}\right)\right)}_{j}\right]$

$$=\left(\frac{E_{0}}{2}\right)^{2}\left[-j\cos\left(\frac{\pi m(t)}{V_{\pi}}\right)+\sin\left(\frac{\pi m(t)}{V_{\pi}}\right)+j\cos\left(\frac{\pi m(t)}{V_{\pi}}\right)+\sin\left(\frac{\pi m(t)}{V_{\pi}}\right)\right]$$

$$=\boxed{\frac{E_{0}^{2}}{2}\sin\left(\frac{\pi m(t)}{V_{\pi}}\right)}\stackrel{?}{=}\frac{E_{0}^{2}}{2}\left(1-\cos\left(\frac{\pi m(t)}{V_{\pi}}\right)\right)$$

If we had sin², then we might have been able to use the half-angle formula. But I am not sure what to do here.

So, how can I get from $\frac{E_{0}^{2}}{2}\sin\left(\frac{\pi m(t)}{V_{\pi}}\right)$ to equation (1)? Did I make a mistake somewhere?

Any help is greatly appreciated.

 

[mfb]

Multiplying by the complex conjugate:

There are sign errors here. What is the complex conjugate of $e^{jx}$?

 

[Charles Link]

Suggestion to make it much simpler: Let $m'=\frac{\pi m}{2V_{\pi}}$. Also write $je^{jx}$ as $e^{j (\pi/2)} e^{jx}$. A few minutes of work including correctly taking complex conjugates should get you the result.

 

[roam]

I see. I made a mistake taking the complex conjugate of the expression. So I used:

$$\overline{\exp\left[j\left(\frac{\pi m}{2V_{\pi}}-\frac{\pi}{4}\right)\right]+j\exp\left[-j\left(\frac{\pi m}{2V_{\pi}}-\frac{\pi}{4}\right)\right]}=\exp\left[-j\left(\frac{\pi m}{2V_{\pi}}-\frac{\pi}{4}\right)\right]-j\exp\left[j\left(\frac{\pi m}{2V_{\pi}}-\frac{\pi}{4}\right)\right]$$

and I got the correct result. Thank you so much for the suggestions.