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Complex number help

Given b = 0.5(a + pi)

Deduce that e^ia - 1 = (2sin(pi/2))e^b

Can anyone give me some help on getting started.
Original post by Roger The Doger
Given b = 0.5(a + pi)

Deduce that e^ia - 1 = (2sin(pi/2))e^b

Can anyone give me some help on getting started.

Hold on, LaTeXing it, because "pi" was very confusing to me (I interpreted it as π\pi, which makes the question trivially nonsense). Answering in a moment. Could you confirm that this is what you mean?

b=12(a+pi)b=\frac{1}{2}(a+p i)
eia−1=2sin⁡(pi2)ebe^{i a}-1 = 2 \sin(\frac{p i}{2}) e^b

ETA: If this is indeed what you mean, I'm not at all sure that the question is answerable. As far as I can tell, it's false. Let p=0, a=1 to see why.
(edited 9 years ago)
Original post by Smaug123
Hold on, LaTeXing it, because "pi" was very confusing to me (I interpreted it as π\pi, which makes the question trivially nonsense). Answering in a moment. Could you confirm that this is what you mean?

b=12(a+pi)b=\frac{1}{2}(a+p i)
eia−1=2sin⁡(pi2)ebe^{i a}-1 = 2 \sin(\frac{p i}{2}) e^b


I'm pretty sure he meant

b=12(a+π)b=\frac{1}{2}(a+\pi)
eia−1=2sin⁡(π2)ebe^{i a}-1 = 2 \sin(\frac{\pi}{2}) e^b

I think it's meant to be easy; possibly to break down into the well known eiπe^{i\pi}? Haven't evaluated it but it looks like it's going to break down pretty nicely from a quick glance.
Original post by anunoriginaluser
I'm pretty sure he meant

b=12(a+π)b=\frac{1}{2}(a+\pi)
eia−1=2sin⁡(π2)ebe^{i a}-1 = 2 \sin(\frac{\pi}{2}) e^b

I think it's meant to be easy; possibly to break down into the well known eiπe^{i\pi}? Haven't evaluated it but it looks like it's going to break down pretty nicely from a quick glance.

But the left-hand side of your final equation is complex, in general, while the right-hand side is real. (Not to mention that sin⁡(π2)\sin(\frac{\pi}{2}) is 1.)
Original post by Smaug123
But the left-hand side of your final equation is complex, in general, while the right-hand side is real. (Not to mention that sin⁡(π2)\sin(\frac{\pi}{2}) is 1.)


Ah yeah; I see your point. I'd jumped to trying to sub a into the second equation, hadn't noticed the 'deduce'. That's a pretty odd question :/
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davros
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Original post by Roger The Doger
Given b = 0.5(a + pi)

Deduce that e^ia - 1 = (2sin(pi/2))e^b

Can anyone give me some help on getting started.


Can you either latex the question or give us a screenshot or a picture?

What you've posted doesn't make a lot of sense in its current form!

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