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Complex functions, (z+i)^5 + (z-i)^5

  • Thread starter L-x
  • Start date
  • #1
L-x
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Homework Statement


solve

(z+i)^5 + (z-i)^5 = 0

Homework Equations





The Attempt at a Solution



I have found the five possible values for z by expanding the original expression into powers of z and i. The imaginary terms cancel exactly, leaving a cubic equation in z which is easily solvable as one of its roots is z=0.
The answers quote the solutions as z = cot[pi(1 + 2n)/10], n = 0, 1, 2, 3, 4. The values I obtained are z=0, z^2=5+sqrt(20), z^2=5-sqrt(20).

My answers are roots of the equation, but the way that the solutions are written suggests that there is a (probably neater) method than the one I used which uses trigonometry. Can anyone see it?
 

Answers and Replies

  • #2
lanedance
Homework Helper
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so you did as follows, to get to a quadratic equation in z^2?
(z+i)^5 = z^5 + 5z^4i + 10z^3i^2 + 10z^2i^3 + 5zi^4 + i^5
(z-i)^5 = z^5 - 5z^4i + 10z^3i^2 - 10z^2i^3 + 5zi^4 - i^5

(z+i)^5 + (z-i)^5 = 2z^5 + 20z^3i^2 + 10zi^4
= 2z^5 - 20z^3 + 10z
= 2z(z^4 - 10z^2 + 5)
 
  • #3
L-x
66
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yes, exactly that.
 
  • #4
lanedance
Homework Helper
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another way to look at it could be
(z+i)^5 = -(z-i)^5
(z+i)^5 = (i-z)^5

but not too sure where to go from here...

might be worth writing cot[pi(1 + 2n)/10] in terms of complex numbers and substituting into the equation to see why it works
 
  • #5
tiny-tim
Science Advisor
Homework Helper
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Hi L-x! :smile:
… the way that the solutions are written suggests that there is a (probably neater) method than the one I used which uses trigonometry. Can anyone see it?
Write the equation as ((z+i)/(z-i))5 = -1.

Then substitute z = cotθ. :wink:
 
  • #6
L-x
66
0
thanks, that's quite a lot neater.
 

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