Solving Fifth Roots of z = 1 + √2

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The discussion focuses on finding the five distinct fifth roots of the complex number z = 1 + √2. The solution involves expressing z in exponential form as z = (1 + √2)e^(2πni), where n takes values from 0 to 4. The fifth roots are derived using the formula z^(1/5) = (1 + √2)^(1/5)(cos(2πn/5) + i sin(2πn/5)). Evaluating this for n = 0, 1, 2, 3, and 4 yields the five distinct roots.

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dogma
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Here's a silly roots question that has my congested mind temporarily stumped:

Let [tex]z = 1 + \sqrt{2}[/tex]. Find the five distinct fifth roots of z.

Thanks in advance for helping me relieve the pressure.
 
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Some of the roots are going to be complex, so the way I would tackle the problem is to rewrite your number in the form:

[tex]z=(1+\sqrt{2}){\rm e}^{2\pi ni}[/tex],

where n=0,1,2,... Then taking the fifth root gives:

[tex]z^{1/5}=(1+\sqrt{2})^{1/5} {\rm e}^{2\pi ni/5}[/tex],

which you can write in the form:

[tex]z^{1/5}=(1+\sqrt{2})^{1/5} \left \{ \cos \left ( \frac{2\pi n}{5} \right) +i \sin \left( \frac{2\pi n}{5} \right) \right \}[/tex].

Evaluating this for different n, should give 5 distinct roots.
 
thank you very much for the insight...I now proceed to kick myself for not seeing it on my own {sound of kicking}

Thanks!
 

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