Help with a factorising step in a proof

[jonel]

Hi all,
I am working through a proof that the addition of two complex number raised to the power of e will obey the normal rules of exponential addition. I have included the actual steps below.

The second step is introducing Euler formula to the imaginary part of the exponent, leaving the real part as it was, outside the bracket.

Step 3 then uses the trigonometric addition formulae as the next step in arriving at the desired result.

Step 4 then shows beautifully the property is obeyed.

The trouble I have is in understanding just how I can get from step 3 to step 4! I have no trouble in going from step 4 to step 3 where simple complex number multiplication leads to step 3.

step 3 is a factorisation problem, but for the life of me I cannot see how to factorise this step.

Any help will be gratefully appreciated.





Thanks

Jonel

 

[Dr.Peterson]

First, this is not "complex number raised to the power of e"; it's e raised to the complex power. You might call it "a complex number exponentiated".

I think you're saying that you see that the steps in question are equal, because you can obtain the first from the second by reversible steps, but you don't see how you would figure out the second if you only had the first.

Set the factor on the left (the exponentials) aside, and focus on the large parenthesized expression. Expand it to a sum of four terms, and then "factor by grouping". It won't look quite right yet; you'll have to change a -1 to i^2 in order to make the two parts have a common factor.

Give that a try, and then show us your work if you get stuck.

 

[jonel]

Thank you for your reply.

Yes, I do agree with your first statement. I specified this incorrectly, I did specify this the wrong way around. But I do understand this part.

The point you mention with regard to expanding to four terms is precisley the point where I became stuck in the first place and led me to at least confirm the step by doing the reverse. I see the sum of two terms (ab-cd) as the real part and (cb-ad) in the imaginary part and that's as far as I can get.

 

[Dr.Peterson]

I was hoping you'd show some actual work so I could see how close you come; but let's just do it:

[MATH](\cos b_1 \cos b_2 - \sin b_1 \sin b_2) + i(\sin b_1 \cos b_2 + \cos b_1 \sin b_2)[/MATH][MATH]=\cos b_1 \cos b_2 - \sin b_1 \sin b_2 + i\sin b_1 \cos b_2 + i\cos b_1 \sin b_2[/MATH] <--- remove parens
[MATH]=\cos b_1 \cos b_2 + i\cos b_1 \sin b_2 - \sin b_1 \sin b_2 + i\sin b_1 \cos b_2[/MATH] <--- rearrange to collect b₁ functions
[MATH]=\cos b_1 (\cos b_2 + i \sin b_2) + \sin b_1 (-\sin b_2 + i \cos b_2)[/MATH] <--- factor each pair; second factors don't match
[MATH]=\cos b_1 (\cos b_2 + i \sin b_2) + i\sin b_1 (i\sin b_2 + \cos b_2)[/MATH] <--- here's the trick! write -1 as i², factor out i
[MATH]=(\cos b_1 + i\sin b_1) (\cos b_2 + i \sin b_2) [/MATH]

 

[jonel]

Thank you very much. That is very helpful. I did try that initial expansion as I thought I might be able to use the the double angle formula, but I got in such a mess that I felt I was not approaching this to the right way. Even as I follow your steps I can see the same mix with a cosb1cosb2 paired with cos b1sinb2 and thought that this was not going to work. It was that 'trick' as you call it that allowed the to parts to be common and extractable as a single term.
I couldn't have come up with step and I have a lot of patience.

Thanks again.

 

[Dr.Peterson]

One thing worth knowing is that this sort of factoring is mostly just the reverse of the multiplication; so the work of factoring can often be reconstructed from the details of the work you already did to check that it is right. But if you don't have any idea of the result, it can be very hard to see what to do -- factoring is an inherently difficult process! And, as in this case, even when you do know the result, it can be hard to do working forward. (I didn't do it without looking at your answer, though I think I could have ...)

 

[jonel]

Thank you. But I now see a recurring theme of your 'trick' with substituting the i² for -1 to get the addition formula to yield the result in a suitable form.