Complex numbers: show that (2+j)e^(1+j3)x + (2-j)e^(1-j3)x is also real

[Naveed]

if x is real show that (2+j)e^(1+j3)x + (2-j)e^(1-j3)x is also real

 

[pka]

if x is real show that (2+j)e^(1+j3)x + (2-j)e^(1-j3)x is also real

This is a mathematics help site so it is customary to use \(\displaystyle \bf{i}\) not \(\displaystyle j\) for the complex unit.
Moreover if by \(\displaystyle j3\) you mean j cubed at least you could type j^3.
\(\displaystyle \bf{i}^n=\begin{cases}\bf{i}& n=1 \\ -1 & n= 2\\-\bf{i} & n=3\\1 & n=4\end{cases}\)
Thus \(\displaystyle e^{(1+\bf{i}^3)}=e^{(1-\bf{i})}\)
Do you know that \(\displaystyle e^{a+b\bf{i}}=e^a\left(\cos(b)+{\bf{i}}\sin(b) \right)~?\)

 

[topsquark]

We need confirmation but "j3" is likely to be \(\displaystyle j \cdot 3\).

Still, the method above is a good way to approach the problem, either way.

-Dan

 

[Dr.Peterson]

if x is real show that (2+j)e^(1+j3)x + (2-j)e^(1-j3)x is also real

To summarize what has been said, for non-physicists the expression would be (adding parentheses that were omitted)

(2+i)e^[(1+3i)x] + (2-i)e^[(1-3i)x]

and the key idea will be to use the fact that e^(a+bi) = e^a(cos(b) + i sin(b)), or in physics notation, e^(a+jb) = e^a(cos(b) + j sin(b)).

The conjugates will result in a lot of canceling, after using the facts that cos(-b) = cos(b) and sin(-b) = -sin(b).