Applying De Moivre's theorem and Euler's formula

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Homework Statement


Text problem: The complex number z satisfies the equation z - z^(-1) = i. Express z^6 in the form a + bi, where a and b are real.

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

The Attempt at a Solution


Handwritten working on lined paper: problem 12, z - z^-1 = i, let z = a+bi, z^-1 = (a-bi)/(a^2+b^2) = z-bar/|z|^2, giving a+bi - (a-bi)/(a^2+b^2) = i, then z^6 = (a+bi)^6 = r^6(cos6θ + i sin6θ) with question marks beside it.


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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DiamondV said:

Homework Statement


Typeset math problem: The complex number z satisfies the equation z - z^(-1) = i. Express z^6 in the form a + bi, where a and b are real.

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

The Attempt at a Solution


Handwritten working on lined paper for problem 12: z - z^-1 = i, letting z = a+bi, rewriting z^-1 as (a-bi)/(a^2+b^2) and then z^6 = (a+bi)^6 = r^6(cos6θ + i sin6θ), with ?? beside the last line.


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##.
 
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##.
Handwritten algebra on lined paper: z - z^(-1) = i, with z = a + bi, then a + bi - 1/(a + bi) = i, multiplied through to a^2 + 2abi - b^2 - 1 = ai - b, and rearranged to a^2 + 2abi - b^2 - 1 - ai + b = 0.

Where do I go form here?
 
DiamondV said:
Handwritten working on lined paper: z - z^(-1) = i, with z = a + bi, expanded to a^2 + 2abi - b^2 - 1 - ai + b = 0.

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?
 
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?
 
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.
 
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?
Handwritten solution to a complex equation z - z^-1 = i with z = a+bi, solving for a and b, then converting z = 1/2 + (1/2)i to polar form with r = √2/2 and θ = cos^-1(1/√2).

Handwritten derivation on lined paper converting z to polar form and applying De Moivre's theorem to get z^6 = (1/8)(cos(3π/2) + i sin(3π/2)), then rectangular form a = 0, b = -1/8, giving z^6 = 0 - (1/8)i.
 
DiamondV said:
Ah. I see now. the entire left side was a complex number. Is this correct now?
Handwritten working on lined paper solving z - z^-1 = i with z = a + bi, splitting into real and imaginary parts, giving a^2 - b^2 - 1 + b = 0 and 2ab - a = 0, then b = 1/2 and a = 1/2, converted to polar form.
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##.
 
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.
 
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.
 
DiamondV said:

Homework Statement


Typeset math problem: The complex number z satisfies the equation z - z^(-1) = i. Express z^6 in the form a + bi, where a and b are real.

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

The Attempt at a Solution


Handwritten working on lined paper: z - z^(-1) = i, let z = a+bi, then z^(-1) = (a-bi)/(a^2+b^2) = z-bar/|z|^2, giving a+bi - (a-bi)/(a^2+b^2) = i, and z^6 = (a+bi)^6 = r^6(cos6θ + isin6θ) with question marks beside it.


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##.
 
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.
 

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