complex

Mar 2010
40
0
Hi,

P(z)=\(\displaystyle z^6-2z^3cos(3\alpha)+1\)
How to show that the solutions of P(z) are: \(\displaystyle a_k=e^{i(-\alpha+\frac{2k\pi}{3}}\) and \(\displaystyle b_k=e^{i(+\alpha+\frac{2k\pi}{3}}\)????
 
May 2010
24
5
Hi,

P(z)=\(\displaystyle z^6-2z^3cos(3\alpha)+1\)
How to show that the solutions of P(z) are: \(\displaystyle a_k=e^{i(-\alpha+\frac{2k\pi}{3}}\) and \(\displaystyle b_k=e^{i(+\alpha+\frac{2k\pi}{3}}\)????
Simplest way is to substitute and check whether \(\displaystyle a_k and b_k give us a zero solution.\)
 
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HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Yes the best way to determine whether or not a given number satisfies an equation is, like ques said, to put the number into the equation and do the calculation!

But this equation isn't that difficult to solve. Let \(\displaystyle u= z^3\) so that your equation becomes the quadratic equation \(\displaystyle u^2- 2ucos(3\alpha)+ 1= 0\). Solve that using the quadratic formula and then Use DeMoivre's formula to solve \(\displaystyle z^3= u\).
 
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