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 Complex Number and Power Puzzle (Posted on 2015-11-23)
Determine all values of θ with 0 ≤ θ ≤ 360o for which the complex number
z = cos θ + i*sin θ satisfies z28 - z8 -1 = 0

 No Solution Yet Submitted by K Sengupta No Rating

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 Solution Comment 1 of 1
If you use the identity z^n = cos(nθ)+isin(nθ) the equation becomes
cos(28θ)+isin(28θ)-cos(8θ)-isin(8θ)-1=0

Separating this into real and imaginary components gives the system:
cos(28θ)-cos(8θ)-1=0
sin(28θ)-sin(8θ)=0

Note that each of these is periodic with period 360/4 = 90 degrees (4 being the GCD of 8 and 28.)
This could be used along with a lot of identities to reduce and solve analytically, but I didn't feel like it.  So I just made a table to get the solutions of each on the interval [0,90)
The cosines equation has 8 solutions and the sine has 12.  The two in common are 15 and 75.

Add the period to give all the solutions sought:
θ={15,75,105,165,195,255,285,345}

 Posted by Jer on 2015-11-26 09:51:41

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