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Roots to Complex Equations

  • Thread starter Moridin
  • Start date
  • #1
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Homework Statement



Determine how many roots the equation

[tex](z + \frac{i\sqrt{3}}{2})^{29} = \frac{1+i}{\sqrt{2}}[/tex]

has that are in the first quadrant.

The Attempt at a Solution



I would like to treat the right hand side in the following way.

[tex](z + \frac{i\sqrt{3}}{2})^{29} = \frac{1+i}{\sqrt{2}}[/tex]

[tex]z + \frac{i\sqrt{3}}{2} = (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})^{1/29} = \cos \frac{\pi}{4 \cdot 29} + i \sin \frac{\pi}{4 \cdot 29}[/tex]

It seems reasonable to rewrite the left hand side into

[tex]z + \frac{i\sqrt{3}}{2} = z + i \sin \frac{\pi}{3}[/tex]

Which give us

[tex]z + i \sin \frac{\pi}{3} = \cos \frac{\pi}{4 \cdot 29} + i \sin \frac{\pi}{4 \cdot 29}[/tex]

Now the real part of the LHS must match the real part of the RHS. This means that the real part of z, must be

[tex]Re ~z~ = \cos \frac{\pi}{4 \cdot 29}[/tex]

and that the imaginary part of x must be

[tex]Im ~z ~= \sin \frac{\pi}{4 \cdot 29} - \sin \frac{\pi}{3}[/tex]

From here, I am pretty much lost.
 

Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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249
Hi Moridin! :smile:

Yes, that all looks pretty good (and nice LaTeX, by the way)! :smile:

Your only error is in:
[tex]z + \frac{i\sqrt{3}}{2} = (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})^{1/29} = \cos \frac{\pi}{4 \cdot 29} + i \sin \frac{\pi}{4 \cdot 29}[/tex]
because (something)^(1/29) has 29 roots.

(much as √(something) = (something)^(1/2) has 2 roots)

So it should be:

[tex]z + \frac{i\sqrt{3}}{2} = (\cos \frac{\pi}{4} + i \sin \frac{\pi}{4})^{1/29} = \cos \left(\frac{\pi}{4 \cdot 29}\,+\,\frac{2n \pi}{29}\right) + i \sin \left(\frac{\pi}{4 \cdot 29}\,+\,\frac{2n \pi}{29}\right)[/tex]

for 0 ≤ n ≤ 28. :smile:
 

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