The Mysteries of De Moivres Theorem and Euler's Formula

In summary: There is no need for DeMoivre or Euler. Once you have ##z = a + ib## you can quite quickly and easily get ##z^2##, then get ##z^4 = z^2 \cdot z^2## and finally ##z^6 = z^4 \cdot z^2##.However, if you do care to use DeMoivre/Euler you should first examine very carefully the actual form of ##z##.Multiplying the equation z+1/z=i by z, you get a quadratic equation for. Get the solution with the quadratic formula. Write z in the exponential/ trig
  • #1
DiamondV
103
0

Homework Statement


0f9c29f81d.png

2. Homework Equations [/B]
De Moivres Theorem/ Eulers formula

The Attempt at a Solution


fa49e98d1c.jpg


Honestly don't know where to go with this now. I already applied De Moivres theorem at the very end. It feels like I have to do something more with either De Moivres theorem or Eulers formula. (only 2 we have done)
 
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  • #2
DiamondV said:

Homework Statement


0f9c29f81d.png

2. Homework Equations [/B]
De Moivres Theorem/ Eulers formula

The Attempt at a Solution


fa49e98d1c.jpg


Honestly don't know where to go with this now. I already applied De Moivres theorem at the very end. It feels like I have to do something more with either De Moivres theorem or Eulers formula. (only 2 we have done)
Start from ##a+bi-\frac{1}{a+bi}=i##, multiply both sides with ##a+bi## and work it out. That will allow you to find the value(s) of a and b. Then use De Moivre's formula to get ##z^6##.
 
  • #3
Samy_A said:
Start from ##a+bi-\frac{1}{a+bi}=i##, multiply both sides with ##a+bi## and work it out. That will allow you to find the value(s) of a and b. Then use De Moivre's formula to get ##z^6##.
f3d25612e5.jpg

Where do I go form here?
 
  • #4
DiamondV said:
f3d25612e5.jpg

Where do I go form here?
On the left you have a complex number, say ##X+iY##, that is equal to 0. What can you then conclude about X and Y?
 
  • #5
Samy_A said:
On the left you have a complex number, say ##X+iY##, that is equal to 0. What can you then conclude about X and Y?
then X and Y should be equal to 0. But I don't see where is the complex number on the left? I know b-ai is one but I'm not too sure about the 2abi thing?
 
  • #6
DiamondV said:
then X and Y should be equal to 0.
Correct.
DiamondV said:
But I don't see where is the complex number on the left? I know b-ai is one but I'm not too sure about the 2abi thing?
The complex number on the left is ##a²+2abi-b²-1-ai+b=(a²-b²-1+b)+(2ab-a)i##. Your computation showed that this complex number is 0. You should be able to get the value(s) of a and b from that.
 
  • #7
Samy_A said:
Correct.

The complex number on the left is ##a²+2abi-b²-1-ai+b=(a²-b²-1+b)+(2ab-a)i##. Your computation showed that this complex number is 0. You should be able to get the value(s) of a and b from that.
Ah. I see now. the entire left side was a complex number. Is this correct now?
d498220a6f.jpg

1e3f200bd8.jpg
 
  • #8
DiamondV said:
Ah. I see now. the entire left side was a complex number. Is this correct now?
d498220a6f.jpg
b is correct.
But you made a sign error while computing a, so that a is not correct.
Keep in mind that an equation in the form ##a²=c## can have two real solutions.

Also, to be complete, you should explain why in ##2ab=a##, you reject the solution ##a=0##.
 
  • #9
Samy_A said:
b is correct.
But you made a sign error while computing a, so that a is not correct.
Keep in mind that an equation in the form ##a²=c## can have two real solutions.

Also, to be complete, you should explain why in ##2ab=a##, you reject the solution ##a=0##.

Ah I see. Should it be a = + or - sqrt(1/4). therefore giving me two solutions for a. a = +1/4 and a = -1/4. I entered both into 2ab-a=0 and both give me true statements.
Also would the reason for rejecting a=0 be that if a=0 it would give me no value for b either.
 
  • #10
DiamondV said:
Ah I see. Should it be a = + or - sqrt(1/4). therefore giving me two solutions for a. a = +1/4 and a = -1/4. I entered both into 2ab-a=0 and both give me true statements.
Any value for a will give you a true statement if b=1/2.
No, that is still wrong. As I wrote, somewhere in the calculation you make a sign error: a minus for no reason becomes a plus.
DiamondV said:
Also would the reason for rejecting a=0 be that if a=0 it would give me no value for b either.
Correct.
 
  • #11
DiamondV said:

Homework Statement


0f9c29f81d.png

2. Homework Equations [/B]
De Moivres Theorem/ Eulers formula

The Attempt at a Solution


fa49e98d1c.jpg


Honestly don't know where to go with this now. I already applied De Moivres theorem at the very end. It feels like I have to do something more with either De Moivres theorem or Eulers formula. (only 2 we have done)

There is no need for DeMoivre or Euler. Once you have ##z = a + ib## you can quite quickly and easily get ##z^2##, then get ##z^4 = z^2 \cdot z^2## and finally ##z^6 = z^4 \cdot z^2##.

However, if you do care to use DeMoivre/Euler you should first examine very carefully the actual form of ##z##.
 
  • #12
Multiplying the equation z+1/z=i by z, you get a quadratic equation for. Get the solution with the quadratic formula. Write z in the exponential/ trigonometric form.
 

Related to The Mysteries of De Moivres Theorem and Euler's Formula

1. What is De Moivre's Theorem and Euler's Formula?

De Moivre's Theorem is a mathematical formula that relates complex numbers to trigonometric functions. It states that for any complex number z and any positive integer n, (cos θ + i sin θ)^n = cos nθ + i sin nθ. Euler's Formula is a special case of De Moivre's Theorem when n = 1, which states that e^(iθ) = cos θ + i sin θ.

2. What are the applications of De Moivre's Theorem and Euler's Formula?

De Moivre's Theorem and Euler's Formula have many applications in mathematics and physics. They are commonly used to simplify calculations involving complex numbers, and they have applications in fields such as electrical engineering, signal processing, and quantum mechanics.

3. How are De Moivre's Theorem and Euler's Formula related?

Euler's Formula is a special case of De Moivre's Theorem, as it can be derived by setting n = 1 in De Moivre's Theorem. Essentially, Euler's Formula is just a simplified version of De Moivre's Theorem.

4. Can De Moivre's Theorem and Euler's Formula be used for negative values of n?

Yes, De Moivre's Theorem and Euler's Formula can be used for negative values of n. In these cases, the formula will result in the reciprocal of the positive n value. For example, if n = -3, then (cos θ + i sin θ)^-3 = (cos θ + i sin θ)^3 = cos 3θ + i sin 3θ.

5. Are there any limitations to De Moivre's Theorem and Euler's Formula?

De Moivre's Theorem and Euler's Formula are limited to calculations involving complex numbers. They cannot be used for real numbers or other types of numbers. Additionally, they have limitations in certain circumstances, such as when n is a non-integer value or when θ is a complex number.

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