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De Moivre's Theorem

Hi Guys,

I've got stuck on question 1b on this problem sheet - see attached pics. I've also uploaded the answers as well.

I see that there should be 4 roots as you have w^2, so youll have a plus and minus version of w which in itself will have two roots, like in question 1a. However from my working out i get ((2^1/2)/2 + (i(5^1/2))/2) as one of them for example so have no idea how they introduced exponential's into there answers.

Any help would be appreciated.

Cheers
20160515_182631.jpg520.3KB
20160515_182613.jpg504.8KB
Original post by trm1
Hi Guys,

I've got stuck on question 1b on this problem sheet - see attached pics. I've also uploaded the answers as well.

I see that there should be 4 roots as you have w^2, so youll have a plus and minus version of w which in itself will have two roots, like in question 1a. However from my working out i get ((2^1/2)/2 + (i(5^1/2))/2) as one of them for example so have no idea how they introduced exponential's into there answers.

Any help would be appreciated.

Cheers


The exponential form of a complex number...
Reply 2
Avatar for trm1
trm1
OP
Original post by zetamcfc
The exponential form of a complex number...


I've not come across that yet...?

After looking it up, are you referring to this:

z=re^i(theta)?
Original post by trm1
I've not come across that yet...?

After looking it up, are you referring to this:

z=re^i(theta)?


Yes, may I ask why you are doing these question if you have not come across this fact?
Reply 4
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trm1
OP
Original post by zetamcfc
Yes, may I ask why you are doing these question if you have not come across this fact?


Dunno, im gunna ask my lecturer' and ill get back to you :wink:

Ohhh but just looking over it i can see where it comes from:

weve gone through e^i(theta) = cos(theta) + i sin(theta) before.

As z = r(cos(theta) + i sin(theta)) - polar form of a complex number

then z = re^i(theta)...

i just didnt spot the link and we've never expressed roots in terms of exponentials before.

Cheers mate
Original post by trm1
Dunno, im gunna ask my lecturer' and ill get back to you :wink:

Ohhh but just looking over it i can see where it comes from:

weve gone through e^i(theta) = cos(theta) + i sin(theta) before.

As z = r(cos(theta) + i sin(theta)) - polar form of a complex number

then z = re^i(theta)...

i just didnt spot the link and we've never expressed roots in terms of exponentials before.

Cheers mate


All good :smile:
Reply 6
Avatar for trm1
trm1
OP
Original post by zetamcfc
All good :smile:


I've attached my working, i seem to have got the first of the four roots correct, but the second one I get 4/3Pi not 2/3Pi, but i cant see where ive gone wrong?
De Moivre's Working.jpg522.7KB
Original post by trm1
I've attached my working, i seem to have got the first of the four roots correct, but the second one I get 4/3Pi not 2/3Pi, but i cant see where ive gone wrong?


4/3 pi = -2/3 pi
Original post by zetamcfc
4/3 pi = -2/3 pi


Whilst I know what you mean this statement hurt my eyes...
Original post by IrrationalRoot
Whilst I know what you mean this statement hurt my eyes...


Easiest way to get the point across.

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