# Convert complex valued sinusoid to real valued sinusoid

This is the homework problem: convert $$x[n]=je^{j\pi n/8}-je^{-j\pi n/8}$$ to a real valued sinusoid.

I understand that $$\sin\theta=\dfrac{e^{j\theta}-e^{-j\theta}}{2j}$$

In the solution, the answers claim that $$x[n]=\dfrac{-e^{j\pi n/8}+e^{-j\pi n/8}}{j}$$, and I don't understand how to get from the original

$$x[n]=je^{j\pi n/8}-je^{-j\pi n/8}$$

and arrive at

$$x[n]=\dfrac{-e^{j\pi n/8}+e^{-j\pi n/8}}{j}$$

after which it is easy to see $$x[n]=-2\sin(\pi n/8)$$

HINT:

It is based on the fact that

$$\sin(x) = \frac{e^{jx} - e^{-jx}}{2j}\quad\text{and}\quad j^2 = -1$$

• Great hint. Multiply the top and bottom by $j$ – Idr Dec 29 '20 at 14:53
• Exactly! Well done. :) – Gilles Dec 29 '20 at 16:53

It's because

$$j \cdot ( a + b) = -\frac{a + b}{j}$$ which stems from the fact that the imaginary unit $$j$$ has the property :

$$j = \frac{-1}{j}$$