Help w/ finding all solns of complex eqns: (z−i)^6=−64, z^2−z^(−2)=4i, z+z ^(−1)=2¯z

[Chicken]

Help w/ finding all solns of complex eqns: (z−i)^6=−64, z^2−z^(−2)=4i, z+z ^(−1)=2¯z

Ive been working on these for days but cant find any notes from my tutorials or lectures on how to solve them need help please!
I got a solutions to the first 2 but am in no way confident in saying they are correct!

a) Find all numbers z ∈ C such that (z − i)^6 = −64.
b)Find all z ∈ C such that z^2 − z^(−2) = 4i.
c) Find all z ∈ C such that z + z ^(−1) = 2¯z .

 

[mmm4444bot]

Ive been working on these for days …

Please show some of that work, so we can see where you're at. Thank you. :cool:

 

[Chicken]

Please show some of that work, so we can see where you're at. Thank you. :cool:

This might be a little messy to read but its what I've got so far. Also the photo of my first answer isnt uploading but i got "-i" and "3i" as the answers.

 

[mmm4444bot]

… cant find any notes from my tutorials or lectures on how to solve …

… the photo of my first answer isnt uploading but i got "-i" and "3i" as the answers.

There are six solutions because you're taking the 6th root of a Complex number.

One way is solve to get:

\(\displaystyle z = i + 2 \cdot \sqrt[6]{-1}\)

where the 6th roots of -1+0∙i are given by:

\(\displaystyle \displaystyle cos\bigg(\frac{(1+2k)\pi}{6}\bigg) + i \cdot sin\bigg(\frac{(1+2k)\pi}{6}\bigg)\)

for k = {0, 1, 2, 3, 4, 5}

---

When k = 1, z = 3i

When k = 4, z = -i

You have yet to work out z when k = {0, 2, 3, 5}

Does any of this look familiar? :cool:

 

[Chicken]

There are six solutions because you're taking the 6th root of a Complex number.

One way is solve to get:

\(\displaystyle z = i + 2 \cdot \sqrt[6]{-1}\)

where the 6th roots of -1+0∙i are given by:

\(\displaystyle \displaystyle cos\bigg(\frac{(1+2k)\pi}{6}\bigg) + i \cdot sin\bigg(\frac{(1+2k)\pi}{6}\bigg)\)

for k = {0, 1, 2, 3, 4, 5}

---

When k = 1, z = 3i

When k = 4, z = -i

You have yet to work out z when k = {0, 2, 3, 5}

Does any of this look familiar? :cool:

Oh yes that helps alot thank you!!!

 

[mmm4444bot]

… yes that helps …

Then you can refresh your memory, by googling keyphrase nth roots of complex numbers

There are a lot of lessons/examples, like this short video. (Note: at [01:40] he goofs up some grouping symbols, but the meaning becomes clear as he finishes the example.) :cool:

 

[HallsofIvy]

Ive been working on these for days but cant find any notes from my tutorials or lectures on how to solve them need help please!
I got a solutions to the first 2 but am in no way confident in saying they are correct!

a) Find all numbers z ∈ C such that (z − i)^6 = −64.

64= 2^6 so one solution is z- i= 2i. The other 5 values for z- i lie equally spaced around the circle |z-i|= 2 in the complex plain.

b)Find all z ∈ C such that z^2 − z^(−2) = 4i.

Multiplying on both sides by z^2, z^4- 1= 4iz^2 so z^4- 4iz^2- 1= 0. Solve for two values for z^2 by completing the square or using the quadratic formula, then find both square roots of each solution. Multiplying by something involving an unknown can introduce "spurious" roots so check all solutions in the original equation.

c) Find all z ∈ C such that z + z ^(−1) = 2¯z .

Multiplying on both sides by z, z^2+ 1= 2zz*. Writing z= x+ iy, x^2- y^2+ 2ixy+ 1= 2x^2- 2y^2. Setting the real parts and imaginary parts on both sides equal, x^2- y^2+ 1= 2x^2- 2y^2, so x^2- y^2- 1= 0 and 2ixy= 0.