The Mysteries of De Moivres Theorem and Euler's Formula

[DiamondV]

Homework Statement

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

The Attempt at a Solution

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)

 

[Samy_A]

Homework Statement

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

The Attempt at a Solution

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 $z^6$.

 

[DiamondV]

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 $z^6$.

Where do I go form here?

 

[Samy_A]

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?

 

[DiamondV]

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?

 

[Samy_A]

then X and Y should be equal to 0.

Correct.

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.

 

[DiamondV]

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?

 

[Samy_A]

Ah. I see now. the entire left side was a complex number. Is this correct now?

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 $a=0$.

 

[DiamondV]

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.

 

[Samy_A]

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.

Also would the reason for rejecting a=0 be that if a=0 it would give me no value for b either.

Correct.

 

[Ray Vickson]

Homework Statement

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

The Attempt at a Solution

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 $z$.

 

[ehild]

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.