# Equating components of complex expansionWatch

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#1
I had a question asking me to find the real and imaginary component of (1+e^iθ)^n. I am confused about why we need to convert this form to (e^iθ/2(e^-iθ/2 + e^iθ/2) and then go on to simply it. Could someone explain why we need to do this?
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9 months ago
#2
because you should know a simple identity involving e^iθ/2 + e^-iθ/2
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#3
(Original post by begbie68)
because you should know a simple identity involving e^iθ/2 + e^-iθ/2
I am struggling to understand why I can’t just equate the real and imaginary component straight away from (1+e^iθ)^n.
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9 months ago
#4
really? you want to expand THAT bracket?
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9 months ago
#5
how you gonna do that?
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#6
(Original post by begbie68)
really? you want to expand THAT bracket?
I was thinking about just taking out real component like (1+cosθ)^n?
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9 months ago
#7
(Original post by Janej77)
I was thinking about just taking out real component like (1+cosθ)^n?
That's not the real component.
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#8
(Original post by DFranklin)
That's not the real component.
So to equate the real and imaginary component, you have to put them in form of x+iy first and can’t just take out the real component when it’s another form? Have I got that right?
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9 months ago
#9
(Original post by DFranklin)
That's not the real component.
yeah, Jane's forgotten about the even powers of isinθ , and it'd seem she's keen to use all those binomial coeffs, too ....
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9 months ago
#10
(Original post by Janej77)
So to equate the real and imaginary component, you have to put them in form of x+iy first and can’t just take out the real component when it’s another form? Have I got that right?
yes, or 1 + cosθ + isinθ ...

use what they gave you, and the identity that comes from it ...
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#11
(Original post by begbie68)
yes, or 1 + cosθ + isinθ ...

use what they gave you, and the identity that comes from it ...
Thanks for your help. That clears it up for me.
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