Convert sinusoidal form to exponential form

Question

How do I get from

$$-2j=\cos\left(\frac{\pi}{2}\right) + 2j\sin\left(\frac{\pi}{2}\right)$$

to the following?

$$-2j=2e^{-j\frac{\pi}{2}}$$

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Background:

Given $x(t)=10+3\cos\omega_0t+5\cos(2\omega_0t+30°)+4\sin3\omega_0t$, its period is $T_0=\frac{2\pi}{\omega_0}$

Using Euler's relation,

$x(t)=10+\frac{3}{2}(e^{j\omega_0t}+e^{-j\omega_0t})+\frac{5}{2}[e^{j(2\omega_0t+30°)}+e^{-j(2\omega_0t+30°)}]+\frac{4}{2j}(e^{j3\omega_0t}-e^{-j3\omega_0t})$

I'd like to know, in detail, how the last term of the equation $\frac{4}{2j}(e^{j3\omega_0t}-e^{-j3\omega_0t})$ can be rearranged to $2e^{-j\frac{\pi}{2}}e^{j3\omega_0t}+2e^{j\frac{\pi}{2}}e^{-j3\omega_0t}$.

Answer 1

You need to observe that $\frac{4}{2j}=-2j=-2e^{j\pi/2}=2e^{-j\pi/2}$, then the rest follows.

The result used is that if $j=e^{j\theta}=\cos(\theta)+j\sin(\theta)$ then $\theta=\pi/2+2n\pi,\mbox{ for }n=0, \pm1,...$,that is you want $\cos(\theta)=0$ and $\sin(\theta)=1$

Answer 2

First of all,

$$-2j\neq\cos(\frac{\pi}{2})+2j\sin(\frac{\pi}{2})$$

But one can write,

$$-2j=\cos(\frac{\pi}{2})-2j\sin(\frac{\pi}{2}) \, \ldots \, Eqn.(1)$$

Adding $\cos(\frac \pi2)$ to both sides of Eqn.(1) and arranging components yields,

$$-2j + \cos(\frac \pi2) =\cos(\frac{\pi}{2}) - j\sin(\frac{\pi}{2}) + \cos(\frac \pi2) - j\sin(\frac{\pi}{2}) \, \ldots \, Eqn.(1)$$

By using the Euler's formula it can be written as,

$$-2j + \cos(\frac \pi2) =2e^{-j\pi/2} \, \ldots \, Eqn.(1)$$

Note:

$\cos(\frac \pi2) = \cos(\frac {-\pi}{2})$

$\sin(\frac \pi2) = -\sin(\frac {-\pi}{2})$

$\cos(\frac \pi2) = 0 $

So,

$$-2j = 2e^{-j\pi/2} \, \ldots \, Eqn.(1)$$

But these steps are unnecessary, complex exponentials has properties of real exponentials, there are shorter ways to show this equality. Also @Conrad Turner explained it before me.