Complex Numbers: Solving Equations with z and zeta?

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SUMMARY

The discussion focuses on solving the equation (z^2 + 1)^4 = 1 for complex numbers z. The solution involves rewriting 1 as e^{ik2\pi} and taking the fourth root, leading to the equation z^2 + 1 = e^{i\pi\frac{k}{2}}. By determining the distinct values of k that yield unique complex numbers, the discussion concludes that there are 8 solutions for z, derived from the 4 solutions of the equation \zeta^4 = 1, where \zeta = z^2 + 1.

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  • Familiarity with Euler's formula e^{ix} = cos(x) + i*sin(x)
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  • Study the properties of complex roots and their geometric interpretations
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lektor
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Hi, I have no clue how to approach this question, was in my last years final exams.

(z^2 + 1)^4 = 1

Find all solutions, where z is a complex number.

Tips please?
 
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First off, write [itex]1=e^{ik2\pi}[/itex]
where k is an integer.
Now, take the fourth root of the equation:
[tex]z^{2}+1=e^{i\pi\frac{k}{2}}[/tex]
For how many choices of k will the right-hand side represent DISTINCT complex numbers?
 
Supposing you have some experience solving equations of the type [itex]z^n=a[/itex] where a is complex...

Set [itex]\zeta = z^2+1[/itex]. Then find all 4 solutions of [itex]\zeta^4 = 1[/itex]. Then go back to [itex]\zeta -1 = z^2[/itex] and for all four [itex]\zeta[/itex] found, find the 2 solutions of z associated, for a total of 8.
 
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