Using Euler's Formula to write a fraction in another form

[WhiteWolf98]

Homework Statement

I'm trying to get from the top fraction to the bottom fraction:
$$u-iv=U_{\infty}\frac{e^{-i\alpha}-e^{i\alpha -2i\theta}+2i\sin(\alpha)e^{-i\theta}}{1-e^{-2i\theta}}$$
$$=U_{\infty}\left[\cos(\alpha)+\sin(\alpha)\frac{1-\cos(\theta)}{\sin(\theta)}\right]$$

Relevant Equations

$$e^{i\theta}=\cos(\theta)+i\sin(\theta)$$
$$e^{-i\theta}=\cos(\theta)-i\sin(\theta)$$
$$e^{2i\theta}=\frac{e^{i\theta}}{e^{-i\theta}}=\frac{\cos(\theta)+i\sin(\theta)}{\cos(\theta)-i\sin(\theta)}$$

Greetings.

I'm having a bit of difficulty with getting from the first to the second equation. I know some basic identities, but it all just feels like a mess. My approach was just going to be to write whatever I could, but some of the terms are confusing me.

$$e^{i\alpha-2i\theta}=\frac{e^{i\alpha}}{e^{2i\theta}}=e^{i\alpha}\div\left(\frac{e^{i\theta}}{e^{-i\theta}}\right)=\cos(\alpha)+i\sin(\alpha)\div\left(\frac{\cos(\theta)+i\sin(\theta)}{\cos(\theta)-i\sin(\theta)}\right)=\frac{(\cos(\alpha)+i\sin(\alpha))(\cos(\theta)-i\sin(\theta))}{\cos(\theta)+i\sin(\theta)}$$

$$e^{-2i\theta}=\frac{1}{e^{2i\theta}}=\frac{e^{-i\theta}}{e^{i\theta}}=\frac{\cos(\theta)-i\sin(\theta)}{\cos(\theta)+i\sin(\theta)}$$

Is this correct...? I think I could probably try again from here. Thank you

 

[suremarc]

Multiplying the numerator and denominator by $e^{i\theta}$ removes $2\theta$ from the equation, which makes converting to the bottom fraction easier. Alternatively, you could expand the exponential terms using Euler’s formula, then use double-angle identities to simplify.

Also, the two equations you gave don’t appear to be equivalent. Is this from a homework problem? When reducing the top equation I got something different from the bottom one.

Edit: ignore the second paragraph, I dropped a minus sign

 

[haruspex]

As @suremarc writes, first step is to multiply top and bottom by $e^{i\theta}$.
Next is to spot that two terms at the top collapse to something involving $\sin(\theta-\alpha)$.

 

[WhiteWolf98]

Hello. Thanks for your replies, much appreciated. I tried again following your advice, and I couldn't eliminate $2θ$. This is what I ended up with:
$$=U_{\infty}\frac{e^{i\theta+i\alpha}-e^{i\alpha-i\theta}+2\sin(\alpha)i}{e^{2i\theta}}$$

Well actually, this was one way of writing it. The other way was:
$$=U_{\infty}\frac{e^{i\theta+i\alpha}-e^{i\alpha}\cdot+\frac{1}{e^{i\theta}}2\sin(\alpha)i}{e^{2i\theta}}$$

I don't think I'm correct... and also, am I supposed to be writing each $e$ so it's on its own? If I can separate out each $e$, then I can write it out in terms of $\sin$ and $\cos$

 

[SammyS]

Hello. Thanks for your replies, much appreciated. I tried again following your advice, and I couldn't eliminate $2θ$. This is what I ended up with:
$$=U_{\infty}\frac{e^{i\theta+i\alpha}-e^{i\alpha-i\theta}+2\sin(\alpha)i}{e^{2i\theta}}$$

How did you get that denominator?

What do you get when you multiply $\displaystyle (1-e^{-2i\theta} ) \cdot e^{i\theta}$ ?

 

[haruspex]

$$=U_{\infty}\frac{e^{i\theta+i\alpha}-e^{i\alpha-i\theta}+2\sin(\alpha)i}{e^{2i\theta}}$$

As @SammyS notes, your denominator is quite wrong, but also have a mistake in the numerator. In the original there was an $e^{-i\alpha}$ which seems to have turned into an $e^{+i\alpha}$.

 

[WhiteWolf98]

How did you get that denominator?

What do you get when you multiply $\displaystyle (1-e^{-2i\theta} ) \cdot e^{i\theta}$ ?

I'll try write out how I got it as soon as I can. For now, when multiplying that out, I get $e^{i\theta}-e^{-i\theta}$

As @SammyS notes, your denominator is quite wrong, but also have a mistake in the numerator. In the original there was an $e^{-i\alpha}$ which seems to have turned into an $e^{+i\alpha}$.

Ah, that is a mistake on my part. Though I don't suppose $e^{i\theta-i\alpha}$ would make it correct

 

[haruspex]

I'll try write out how I got it as soon as I can. For now, when multiplying that out, I get $e^{i\theta}-e^{-i\theta}$
Ah, that is a mistake on my part. Though I don't suppose $e^{i\theta-i\alpha}$ would make it correct

What's a way of writing $e^{ix}-e^{-ix}$ with trig functions?
Can you see how to get the $e^{i\theta-i\alpha}$ and the one like it into the same form?

 

[WhiteWolf98]

This is just not making any sense, sorry. I'll put all the steps I've done below.

I feel like I'm just going in circles. And then for:

What's a way of writing $e^{ix}-e^{-ix}$ with trig functions?
Can you see how to get the $e^{i\theta-i\alpha}$ and the one like it into the same form?

Depending on which way I do it, I get different answers... which is weird! I don't get it.

Yeah, I don't know anymore, nothing is working

 

[haruspex]

This is just not making any sense, sorry. I'll put all the steps I've done below.

In your first block above, I do not understand how in the denominator you convert $e^{i\theta}-e^{-i\theta}$ to $e^{2i\theta}$.
In your attempt to convert it to a trig function instead, second block, both attempts are correct. The final forms you have are equivalent, but using 2 sin x cos x = sin(2x) in the second attempt was not helpful. Use the result from the first attempt to replace $e^{i\theta}-e^{-i\theta}$ in the denominator.

The first two terms in the numerator form a pair of exactly the same pattern. Can you see that? So you can use the corresponding trig substitution.

 

[WhiteWolf98]

In your first block above, I do not understand how in the denominator you convert $e^{i\theta}-e^{-i\theta}$ to $e^{2i\theta}$.

Ohh, I see it, I see it. What a silly mistake, I can't just add them. Alright, I'll give it another go. Thank you

 

[WhiteWolf98]

Okay, I think I was able to do it. For the sake of completion, I'll put the rest of my solution here.

I realize the substitution probably wasn't necessary, but I found it easier to do thinking about it that way.

Thank you very much everyone for all your help :)

 

[haruspex]

Okay, I think I was able to do it. For the sake of completion, I'll put the rest of my solution here.

View attachment 276798

I realize the substitution probably wasn't necessary, but I found it easier to do thinking about it that way.

Thank you very much everyone for all your help :)

Well done