Exponential form of complex numbers

[jonnburton]

I have been looking at the solution to a question set in my book and have been unable to follow the working beyond a certain point. Can anyone tell me how this works?

If x is real, show that $\displaystyle (2+j)e^{(1+j3)x} + (2-j)e^{(1-j3)x}$ is also real.

Although I wasn't able to do it myself, I understand the first few steps:

$\displaystyle (2+j)e^xe^{(j3x)} + (2-j)e^xe^{-j3x}$


Factor out $\displaystyle e^x$:

$\displaystyle e^x \left[2e^{j3x}+je^{j3x} + 2e^{-j3x}-je^{-j3x}\right]$





Group terms:

$\displaystyle e^x\left[2 (e^{j3x}+e^{-j3x})+j(e^{j3x}-e^{-j3x})\right]$



But I can't see how this next step follows from what we had previously:

$\displaystyle e^x\left[ 4(\frac{e^{j3x}+e^{-j3x}}{2}) -2 (\frac{e^{j3x}-e^{-j3x}}{2j})\right]$



I can see how the left hand part inside the square brackets works: 4/2 =2, so this hasn't changed. But I do not see how the right hand part works, ie $\displaystyle +j(e^{j3x}-e^{-j3x})$ becomes $\displaystyle -2 (\frac{e^{j3x}-e^{-j3x}}{2j})$

​Can anybody tell me how this comes about?

 

[DrPhil]

I have been looking at the solution to a question set in my book and have been unable to follow the working beyond a certain point. Can anyone tell me how this works?

If x is real, show that $\displaystyle (2+j)e^{(1+j3)x} + (2-j)e^{(1-j3)x}$ is also real.

Although I wasn't able to do it myself, I understand the first few steps:

$\displaystyle (2+j)e^xe^{(j3x)} + (2-j)e^xe^{-j3x}$


Factor out $\displaystyle e^x$:

$\displaystyle e^x \left[2e^{j3x}+je^{j3x} + 2e^{-j3x}-je^{-j3x}\right]$





Group terms:

$\displaystyle e^x\left[2 (e^{j3x}+e^{-j3x})+j(e^{j3x}-e^{-j3x})\right]$



But I can't see how this next step follows from what we had previously:

$\displaystyle e^x\left[ 4(\frac{e^{j3x}+e^{-j3x}}{2}) -2 (\frac{e^{j3x}-e^{-j3x}}{2j})\right]$



I can see how the left hand part inside the square brackets works: 4/2 =2, so this hasn't changed. But I do not see how the right hand part works, ie $\displaystyle +j(e^{j3x}-e^{-j3x})$ becomes $\displaystyle -2 (\frac{e^{j3x}-e^{-j3x}}{2j})$

​Can anybody tell me how this comes about?

It works because $\displaystyle 1/j = -j$.

Consider the coefficient $\displaystyle j$ for the second term. Multiply by $\displaystyle (2j)/(2j)$:

$\displaystyle \displaystyle j\ \dfrac{2j}{2j} = \dfrac{2j^2}{2j} = -\dfrac{2}{2j}$

 

[jonnburton]

It works because $\displaystyle 1/j = -j$.

Consider the coefficient $\displaystyle j$ for the second term. Multiply by $\displaystyle (2j)/(2j)$:

$\displaystyle \displaystyle j\ \dfrac{2j}{2j} = \dfrac{2j^2}{2j} = -\dfrac{2}{2j}$

OK, that makes sense now! Thank you DrPhil!