How to write the complex exponential in terms of sine/cosine?

[Selling Papayas]

I apologize in advance if any formatting is weird; this is my first time posting. If I am breaking any rules with the formatting or if I am not providing enough detail or if I am in the wrong sub-forum, please let me know.

1. Homework Statement
Using Euler's formula : e^jx = cos(x) + jsin(x) and the exponential representations of sin & cos, which are derived from Euler's formula:

$\displaystyle \sin(x)=\frac {e^{jx} - e^{-jx}} {2j}$
$\displaystyle \cos(x)=\frac {e^{jx} + e^{-jx}} {2}$
​

write the following in terms of sin & cos:

1. $\displaystyle je^{j \frac {π} {4}t} - je^{-j \frac {π} {4}t} + 2e^{j \frac {5π} {4}t} + 2e^{-j \frac {5π} {4}t}$
2. $\displaystyle \frac {3} {4jkπ} (1-e^{-jkπ})^2$

Homework Equations

I posted the ones given in the problem in part 1, and the only other one I used is $j = e^{j \frac {π} {2}}$
This is probably obvious, but my classes use j instead of i to represent the imaginary number.

The Attempt at a Solution

So for part 1, I think I solved it counter-intuitively (Wolfram Alpha says my solution is equivalent but its extremely long and I still feel as if I'm missing some key conversion)
I converted the j's into $e^j\frac{π}{2}$, multiplied the exponentials together (adding the exponents in each term) and ended up with the following:

$[\cos(\frac {π} {2} + \frac {π} {4}t) + j\sin(\frac {π} {2} + \frac {π} {4}t)] - [\cos(\frac {π} {2} - \frac {π} {4}t) + j\sin(\frac {π} {2} - \frac {π} {4}t)] + 2[\cos(\frac {5π} {4}t) + j\sin(\frac {5π} {4}t)] + 2[\cos(\frac {-5π} {4}t) + j\sin(\frac {-5π} {4}t)]$​

As you can see, it's a lot, but I am confident I at least have a right solution.

For part 2, I am getting stumped pretty quickly. I've tried to apply the methods I used in part 1 to solve this, but can't seem to pull out the k and the π from the denominator.
What I've done so far:

$\displaystyle \frac {3} {4jkπ} (1-e^{-jkπ})^2$
$\displaystyle= \frac {3} {4jkπ} (1-e^{-jkπ})(1-e^{-jkπ})$
$\displaystyle= \frac {3} {4jkπ} (1 - 2e^{-jkπ} + e^{-2jkπ})$
$\displaystyle= \frac {3} {4jkπ} - \frac {3e^{-jkπ}} {2jkπ} + \frac{3e^{-2jkπ}} {4jkπ}$​

 

[LCKurtz]

Please fix your Latex to make your post readable:
https://www.physicsforums.com/help/latexhelp/

 

[Selling Papayas]

Please fix your Latex to make your post readable:
https://www.physicsforums.com/help/latexhelp/

My apologies. I am trying to do that right now. It doesn't seem to be doing fractions like I want it to.

 

[fresh_42]

I fixed it. Sorry, took a while.

 

[Selling Papayas]

I fixed it. Sorry, took a while.

Oh I was fixing it too. I think we were clashing; my bad! I thought I was going crazy or saving it wrong.

 

[Ray Vickson]

I apologize in advance if any formatting is weird; this is my first time posting. If I am breaking any rules with the formatting or if I am not providing enough detail or if I am in the wrong sub-forum, please let me know.

1. Homework Statement
Using Euler's formula : e^jx = cos(x) + jsin(x) and the exponential representations of sin & cos, which are derived from Euler's formula:

$\displaystyle sin(x)=\frac {e^{jx} - e^{-jx}} {2j}$
$\displaystyle cos(x)=\frac {e^{jx} + e^{-jx}} {2}$
​

write the following in terms of sin & cos:

1. $\displaystyle je^{j \frac {π} {4}t} - je^{-j \frac {π} {4}t} + 2e^{j \frac {5π} {4}t} + 2e^{-j \frac {5π} {4}t}$
2. $\displaystyle \frac {3} {4jkπ} (1-e^{-jkπ})^2$

Homework Equations

I posted the ones given in the problem in part 1, and the only other one I used is $j = e^{j \frac {π} {2}}$
This is probably obvious, but my classes use j instead of i to represent the imaginary number.

The Attempt at a Solution

So for part 1, I think I solved it counter-intuitively (Wolfram Alpha says my solution is equivalent but its extremely long and I still feel as if I'm missing some key conversion)
I converted the j's into $e^j\frac{π}{2}$, multiplied the exponentials together (adding the exponents in each term) and ended up with the following:

$[cos(\frac {π} {2} + \frac {pi} {4}t) + jsin(\frac {π} {2} + \frac {pi} {4}t)] - [cos(\frac {π} {2} - \frac {pi} {4}t) + jsin(\frac {π} {2} - \frac {pi} {4}t)] + 2[cos(\frac {5π} {4}t) + jsin(\frac {5π} {4}t)] + 2[cos(\frac {-5π} {4}t) + jsin(\frac {-5π} {4}t)]$​

As you can see, it's a lot, but I am confident I at least have a right solution.

For part 2, I am getting stumped pretty quickly. I've tried to apply the methods I used in part 1 to solve this, but can't seem to pull out the k and the π from the denominator.
What I've done so far:

$\displaystyle \frac {3} {4jkπ} (1-e^{-jkπ})^2$
$\displaystyle= \frac {3} {4jkπ} (1-e^{-jkπ})(1-e^{-jkπ})$
$\displaystyle= \frac {3} {4jkπ} (1 - 2e^{-jkπ} + e^{-2jkπ})$
$\displaystyle= \frac {3} {4jkπ} - \frac {3e^{-jkπ}} {2jkπ} + \frac{3e^{-2jkπ}} {4jkπ}$​

For (a): it is much, much easier to NOT expand out the $j$ as $\exp(j \pi/2)$. Assuming you intend $t$ to be real, we have:
$$\text{your function} = j \times 2j \sin(\pi t/4) + 2 \times 2 \cos(5 \pi t/4).$$
That simplifies even more. Of course, it becomes more complicated if $t$ is complex.

For (b): again, assuming $k$ is real, you can either express $\exp(-j k \pi)$ in terms of $\sin(k \pi), \cos(k \pi)$ then square it all out to get an answer in terms of $\sin(k \pi), \cos(k \pi), \sin^2(k \pi), \cos^2(k \pi)$, or else expand out the $(1-\exp(-j k \pi) )^2$ first, then get an answer in terms of $\sin(k \pi), \cos(k \pi), \sin(2 k \pi), \cos(2 k \pi)$. The choice is up to you, and both are equally valid. Of course the $k$ and $\pi$ cannot be taken out of the denominator; there is no reason to suppose they can or should be taken out.

Again, everything is much more complicated if $k$ is not real.

BTW: your expressions will look much better if you use the LaTeX standard of typing "\sin" instead of "sin" and "\cos" instead of "cos". Here is the difference: $\sin \theta$ instead of $sin \theta$ and $\cos \theta$ instead of $cos \theta$. TeX/LaTeX does not recognize "cos" as a standard mathematical function, it just thinks it is a "c" followed by an "o" followed by an "s".

 

[Selling Papayas]

For (a): it is much, much easier to NOT expand out the $j$ as $\exp(j \pi/2)$. Assuming you intend $t$ to be real, we have:
$$\text{your function} = j \times 2j \sin(\pi t/4) + 2 \times 2 \cos(5 \pi t/4).$$
That simplifies even more. Of course, it becomes more complicated if $t$ is complex.

For (b): again, assuming $k$ is real, you can either express $\exp(-j k \pi)$ in terms of $\sin(k \pi), \cos(k \pi)$ then square it all out to get an answer in terms of $\sin(k \pi), \cos(k \pi), \sin^2(k \pi), \cos^2(k \pi)$, or else expand out the $(1-\exp(-j k \pi) )^2$ first, then get an answer in terms of $\sin(k \pi), \cos(k \pi), \sin(2 k \pi), \cos(2 k \pi)$. The choice is up to you, and both are equally valid. Of course the $k$ and $\pi$ cannot be taken out of the denominator; there is no reason to suppose they can or should be taken out.

Again, everything is much more complicated if $k$ is not real.

BTW: your expressions will look much better if you use the LaTeX standard of typing "\sin" instead of "sin" and "\cos" instead of "cos". Here is the difference: $\sin \theta$ instead of $sin \theta$ and $\cos \theta$ instead of $cos \theta$. TeX/LaTeX does not recognize "cos" as a standard mathematical function, it just thinks it is a "c" followed by an "o" followed by an "s".

Hey thank you so much for the quick response! I used your advice and got a way cleaner answer for #1 (basically the same thing as you, I just distributed the j and the 2). As far as I know, we can assume t and k is real (not sure if they have to be integers or not). For #2, I'm still not sure if I have a simple enough answer. I think I understand what you told me, but is there just not a way to convert every j or coefficient into a type of sine or cosine? I'm getting:

$\displaystyle \frac {3} {4jπk} - \frac {3 \cos(πk)} {2jπk} + \frac {3 \sin(πk)} {2kπ} + \frac {3 \cos(2kπ)} {4jkπ} - \frac {3 \sin(2kπ)} {4πk}$​

 

[Ray Vickson]

Hey thank you so much for the quick response! I used your advice and got a way cleaner answer for #1 (basically the same thing as you, I just distributed the j and the 2). As far as I know, we can assume t and k is real (not sure if they have to be integers or not). For #2, I'm still not sure if I have a simple enough answer. I think I understand what you told me, but is there just not a way to convert every j or coefficient into a type of sine or cosine? I'm getting:

$\displaystyle \frac {3} {4jπk} - \frac {3 \cos(πk)} {2jπk} + \frac {3 \sin(πk)} {2kπ} + \frac {3 \cos(2kπ)} {4jkπ} - \frac {3 \sin(2kπ)} {4πk}$​

One final refinement: it is usual to write a complex number as $a + j b$, so if you have it in the form $c + \frac{d}{j}$ you should convert it (to put the "$j$" in the numerator instead of the denominator). Writing $c + \frac{d}{j}$ is not wrong, but it does go against the world-wide standard.