Solving De Moivre's Theorem: Help Needed

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In summary, the conversation is about the individual's struggle with solving a particular question using de Moivre's theorem and how they keep getting the wrong answer despite checking multiple times. They ask for help in simplifying the expression and writing it in a complex number form. Eventually, it is revealed that the individual made a mistake in inputting the values and the correct answer is 1. The conversation also includes a discussion on the use of "i" and "iota" as the imaginary unit.
  • #1
sarah786
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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...
 
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  • #2
Isn't it already seperated?

Real part = -1/2
Imaginary part = root(3)/2
 
  • #3
no, i have to remove that power "3" somehow... i have to write a complex number to the power 1...
 
  • #4
i have to "simplify" it and write the "simplified form" as a complex number...
 
  • #5
From the looks of it, they want you to go the other direction and write it in a form like [tex]a e^{i \theta}[/tex], is that what the question is asking for?
 
  • #6
olivermsun said:
From the looks of it, they want you to go the other direction and write it in a form like [tex]a e^{i \theta}[/tex], is that what the question is asking for?

no... i think i have very clearly stated my question...
 
  • #7
Please write the expression in a different way. I don't see any power of three.

If (sq.root3) means [tex]\sqrt{3}[/tex] then it's already answered. If it means something else, then it's not obvious.
 
  • #8
oopsss... sorryy... yeahh thanks for telling ... the question is:
[-(1/2) + iota*(sq.root3)/2]^3
 
  • #9
and yea... it means square root of 3
 
  • #10
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.
 
  • #11
By the way, in the standard English, at any rate, the imaginary unit is "i" not "iota".
[itex]1/2+ i\sqrt{3}/2[/itex] in polar form is [itex]cos(\pi/3)+ i sin(\pi/3)[/itex].
Its third power is given by [itex]cos(3(\pi/3))+ i sin(3(\pi/3))= cos(\pi)+ i sin(\pi)= -1[/itex]

As Unrest says, perhaps you made the same mistake I did, and used [itex]1/2+ i\sqrt{3}{2}[/itex] instead of [itex]-1/2+ i\sqrt{3}/2[/itex]!
 
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  • #12
HallsofIvy said:
[itex]cos(\pi)+ i sin(\pi)= 1[/itex]

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

Maybe sarah did the same as you! ;)
 
  • #13
Unrest said:
Maybe sarah did the same as you! ;)

Yes.. i did this same mistake :smile:
 

FAQ: Solving De Moivre's Theorem: Help Needed

1. What is De Moivre's Theorem?

De Moivre's Theorem is a mathematical theorem that allows us to raise a complex number to a power. It states that for any complex number z and any positive integer n, (cos z + i sin z)^n = cos(nz) + i sin(nz).

2. How do I solve De Moivre's Theorem?

To solve De Moivre's Theorem, you can use the formula (cos z + i sin z)^n = cos(nz) + i sin(nz). Plug in the values for z and n, and then use the trigonometric identities to simplify the resulting expression.

3. What are the applications of De Moivre's Theorem?

De Moivre's Theorem has various applications in mathematics, physics, and engineering. It is used to simplify complex calculations involving powers of complex numbers, to solve differential equations, and to understand the behavior of waves and oscillations.

4. Can De Moivre's Theorem be used for all complex numbers?

Yes, De Moivre's Theorem can be used for all complex numbers. However, it is most commonly used for numbers in polar form (r(cosθ + i sinθ)), where r is the magnitude and θ is the angle of the complex number.

5. Are there any alternative methods for solving De Moivre's Theorem?

Yes, there are alternative methods for solving De Moivre's Theorem, such as using the binomial theorem or using the exponential form of complex numbers. However, De Moivre's Theorem is often the most efficient and straightforward method for solving complex number powers.

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