Solving De Moivre's Theorem: Help Needed

[sarah786]

I have practiced solving de moivre's theorem... but i don't know why i keep getting the wrong answer in a particular question although i have checked a hundred times and havn't found a mistake...
this is the question i am talking about:
Simplify the following ( separate into real and imaginary parts ):
[-(1/2) + iota*(sq.root3)/2]

I have tried a hundred times using de moivre's theorem and i get "-1" each time... but when i do the same question using the formula (a+b)^3, i get "1" ... moreover, i have got a solution lest and there it says the answer should be "1" ... please if you can do the steps for me, i'll be very thankful and it'll be a big help... i will then be able to check where i am wrong... thank you...

 

[Unrest]

Isn't it already separated?

Real part = -1/2
Imaginary part = root(3)/2

 

[sarah786]

no, i have to remove that power "3" somehow... i have to write a complex number to the power 1...

 

[sarah786]

i have to "simplify" it and write the "simplified form" as a complex number...

 

[olivermsun]

From the looks of it, they want you to go the other direction and write it in a form like $$a e^{i \theta}$$, is that what the question is asking for?

 

[sarah786]

From the looks of it, they want you to go the other direction and write it in a form like $$a e^{i \theta}$$, is that what the question is asking for?

no... i think i have very clearly stated my question...

 

[Unrest]

Please write the expression in a different way. I don't see any power of three.

If (sq.root3) means $$\sqrt{3}$$ then it's already answered. If it means something else, then it's not obvious.

 

[sarah786]

oopsss... sorryy... yeahh thanks for telling ... the question is:
[-(1/2) + iota*(sq.root3)/2]^3

 

[sarah786]

and yea... it means square root of 3

 

[Unrest]

de Moivre's theorum:
[cos(x) + i sin(x)]ⁿ = cos(nx) + i sin(nx)

n=3
cos(x) = -1/2
sin(x) = sqroot(3)/2
x=120deg

the RHS is
cos(3*120deg) + i sin(3*120deg) = 1

So 1 looks OK.

 

[HallsofIvy]

By the way, in the standard English, at any rate, the imaginary unit is "i" not "iota".
$1/2+ i\sqrt{3}/2$ in polar form is $cos(\pi/3)+ i sin(\pi/3)$.
Its third power is given by $cos(3(\pi/3))+ i sin(3(\pi/3))= cos(\pi)+ i sin(\pi)= -1$

As Unrest says, perhaps you made the same mistake I did, and used $1/2+ i\sqrt{3}{2}$ instead of $-1/2+ i\sqrt{3}/2$!

 

[Unrest]

$cos(\pi)+ i sin(\pi)= 1$

Now, as to why you are getting "-1"..

Maybe sarah did the same as you! ;)

 

[sarah786]

Maybe sarah did the same as you! ;)

Yes.. i did this same mistake