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

• Selling Papayas
In summary: For both parts, I suggest you look carefully at the trig identities for complex arguments, like ##\sin(\theta) = \frac{e^{j \theta} - e^{-j \theta}}{2j}## and see what you can do with these.
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 : ejx = 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}##​

Last edited:
LCKurtz said:
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.

I fixed it. Sorry, took a while.

fresh_42 said:
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.

Selling Papayas said:
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 : ejx = 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".

Last edited:
Delta2 and Selling Papayas
Ray Vickson said:
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}##​

Last edited:
Selling Papayas said:
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.

Selling Papayas

## 1. How do you write the complex exponential in terms of sine and cosine?

To write the complex exponential in terms of sine and cosine, you can use Euler's formula, which states that e^(ix) = cos(x) + i*sin(x). This means that any complex number in the form a + bi can be written as a*cos(b) + i*a*sin(b).

## 2. Can you provide an example of writing a complex exponential in terms of sine and cosine?

Yes, for example, if we have the complex number 3 + 2i, we can write it as 3*cos(2) + i*3*sin(2).

## 3. Why is it useful to write the complex exponential in terms of sine and cosine?

Writing the complex exponential in terms of sine and cosine can make it easier to perform calculations and manipulate complex numbers. It also allows for a better understanding of the relationships between different types of functions.

## 4. Are there any other ways to write the complex exponential?

Yes, there are other ways to write the complex exponential, such as using the hyperbolic functions sinh and cosh, or using the polar form of a complex number. However, using sine and cosine is a commonly used and convenient method.

## 5. Can the complex exponential be written in terms of other trigonometric functions?

Yes, the complex exponential can also be written in terms of the tangent and cotangent functions. Using Euler's formula, e^(ix) = cos(x) + i*sin(x), we can also write it as e^(ix) = (1/2)(cot(x) - i*csc(x)) + (1/2)(tan(x) + i*sec(x)).

### Similar threads

• Calculus and Beyond Homework Help
Replies
1
Views
359
• Calculus and Beyond Homework Help
Replies
11
Views
500
• Calculus and Beyond Homework Help
Replies
3
Views
560
• Calculus and Beyond Homework Help
Replies
2
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
640
• Calculus and Beyond Homework Help
Replies
16
Views
1K
• Calculus and Beyond Homework Help
Replies
6
Views
1K
• Calculus and Beyond Homework Help
Replies
8
Views
1K
• Engineering and Comp Sci Homework Help
Replies
3
Views
236
• Calculus and Beyond Homework Help
Replies
10
Views
1K