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If z is a complex number satisfying z+(1/z)=1, calculate z^10+ (1/z^10).

 Aug 24, 2020
 #1
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You should check out de moivre's theorem

 Aug 24, 2020
 #2
avatar+497 
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since i haven't seen anyone answer yet, i'll try pitching my 2 cents.

we have that z+1/z = 1

if we square our given expression, we get:

 

z^2 + 2 + 1/z^2 = 1

z^2 + 1/z^2 = -1

if we square this again, we get:

z^4 + 2 + 1/z^4 = 1

z^4 + 1/z^4 = -1

as you might be able to deduce, this will equal -1 however many times we square our expression, because the 2ab term will always equal 2, and the RHS(right hand side) will always equal (-1)^2 = 1, so subtract the two, and you get -1.

 

z^2 + 1/z^2 = -1

z^4 + 1/z^4 = -1

z^8 + 1/z^8 = -1

Now we have a problem because we are asked to find z^10 + 1/z^10 which does not entail these powers of 2. We can try experimenting to get the z^10 term, and the first thing that comes to mind is multiplying by z^2 + 1/z^2

(z^2+1/z^2)(z^8+1/z^8) = z^10 + 1/z^10 + z^6 + 1/z^6 = 1

now we need the value of z^6 + 1/z^6 to get the value of our desired expression.

To do so, we can repeat these steps with z^2+1/z^2 and z^4 + 1/z^4

(z^2 + 1/z^2)(z^4+1/z^4) = z^6 + 1/z^6 + z^2 + 1/z^2 = 1

this is good, because we have the value of z^2 + 1/z^2 = -1

z^6 + 1/z^6 - 1 = 1

z^6 + 1/z^6 = 2

now we go back to our expression with the z^10 term and substitute the value of z^6 + 1/z^6 to get;

z^10 + 1/z^10 + 2 = 1

z^10 + 1/z^10 = -1

 Aug 24, 2020

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