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# Complex Numbers watch

1. Hi i was wondering if you could show me how to work out the following please.

Find all solutions to the following equations, expressing your answers in Cartesian form.

(z+1)^4 + (z-1)^4 = 0

thanks.
2. First verify that 1 isn't a solution (so that dividing by z-1 is okay). Then let w = (z+1)/(z-1). See where you get with that.
3. (Original post by generalebriety)
First verify that 1 isn't a solution (so that dividing by z-1 is okay). Then let w = (z+1)/(z-1). See where you get with that.
im not sure what you mean by 'let w = (z+1)/(z-1)'.. can you explain please?
4. (Original post by stats10)
im not sure what you mean by 'let w = (z+1)/(z-1)'.. can you explain please?
I'm making a substitution to make the equation easier to solve. If you've checked that z = 1 isn't a solution, then you're safe to divide by z-1 a few times. Do that, and you'll notice that if you make the substitution I suggested above, your equation becomes a much easier equation in w. Solve for w.
5. Surely it'd just be easier to expand out and take advantage of the fact that some of the terms cancel nicely? probably not. wait, it might be easier, I'm not sure.
6. (Original post by stats10)
im not sure what you mean by 'let w = (z+1)/(z-1)'.. can you explain please?
if 1 isn't a solution then you can divide by (z-1)^4, so you're left with

if then you can change it to

7. (Original post by tommm)
Surely it'd just be easier to expand out and take advantage of the fact that some of the terms cancel nicely?
*shrugs* That works too. Probably is slightly easier, actually. Not by much though.
8. (Original post by generalebriety)
*shrugs* That works too. Probably is slightly easier, actually. Not by much though.
i get: +/- root 3 +/- 4th-root 8.. is this correct?
9. Alternatively dividing by z^4 and making use of some trig is a possibility [though all these methods are pretty quick tbh].

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