Help: Maths, Imaginary numbers/Polar form

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Starburn
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#1
Report Thread starter 16 years ago
#1
If I have e.g. (-1 + i)^7, can I turn -1 + i into polar form (i.e. Root2 e^i(3pi/4) ) then use normal rules to deal with the ^7 from the start? (i.e. multiply the 3pi/4 by 7)? :confused:

Its the quickest way I can see to turn (<Im number>)^<real> into polar form
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JimBob
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#2
Report 16 years ago
#2
(Original post by Starburn)
If I have e.g. (-1 + i)^7, can I turn -1 + i into polar form (i.e. Root2 e^i(3pi/4) ) then use normal rules to deal with the ^7 from the start? (i.e. multiply the 3pi/4 by 7)? :confused:

Its the quickest way I can see to turn (<Im number>)^<real> into polar form
I think so, that would prob be the method I would choose. You could also use the z = cosP + isinP method... using the (cosP + isinP)^n = cos(nP) + isin(nP) identity... but think that would be a lot longer than necessary.

So yeah, I prefer your first method.
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hornblower
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#3
Report 16 years ago
#3
(Original post by JimBob)
I think so, that would prob be the method I would choose. You could also use the z = cosP + isinP method... using the (cosP + isinP)^n = cos(nP) + isin(nP) identity... but think that would be a lot longer than necessary.

So yeah, I prefer your first method.
You don't need De Moivre's theorem do you?
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Starburn
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#4
Report Thread starter 16 years ago
#4
good good

didn't want to write out all the working only to find it was incorrect lol.
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