Find $z^5 +\frac{1}{z^5}$ given that $z > 0$ and $z^2 + \frac{1}{z^2}=14$

Question

I have been struggling with this problem and I think I am nearly there. Here is my working so far:

Any hints?

Answer 1

As $z>0, z+\dfrac1z=4$

There is a typo in your $$\left(z+\dfrac1z\right)^3$$

The right hand side should be $$z^3+\dfrac1{z^3}+3\left(z+\dfrac1z\right)$$

which will give us the value of $z^3+\dfrac1{z^3}$

Finally,

$$\left(z^2+\dfrac1{z^2}\right)\left(z^3+\dfrac1{z^3}\right)=z^5+\dfrac1{z^5}+\left(z+\dfrac1{z}\right)$$

Replace the values of $z+\dfrac1{z},z^2+\dfrac1{z^2}, z^3+\dfrac1{z^3}$

See also : Proving that $\frac{\phi^{400}+1}{\phi^{200}}$ is an integer.

Answer 2

Heres a trick:

1) $(z + \frac 1z)(z + \frac 1z) =$

$z^2 + 1 + 1 + \frac 1z^2 =$

$(z^2 + \frac 1{z^2}) + 2$.

2)$(z+\frac 1z)(z^2 + \frac 1{z^2}) = $

$(z^3 + \frac 1{z^3}) + (z + \frac 1z)$

3) $(z+\frac 1z)(z^3 + \frac 1{z^3}) = $

$(z^4 + \frac 1{z^4})+ (z^2 + \frac 1{z^2})$

4) $(z + \frac 1z)(z^4 + \frac 1{z^4}) = $

$(z^5 + \frac 1{z^5}) + (z^3 + \frac 1z^3)$.

....

From 1) we get

$(z+\frac 1z)^2 = (z^2 + \frac 1{z^2}) + 2 = 14 + 2 = 16$

So $(z+\frac 1z) = 4$.

From 2) we get

$z^3 + \frac 1{z^3} = 4*14 - 4 = 52$.

From 3) we get

$z^4 + \frac 1{z^4} = 4*52 - 14=194$

And from 5) we get

$z^5 + \frac 1{z^5} = 4*194 - 52 = 724$

Answer 3

We already know about $\ z+\frac 1z.\ $ It's also easy to see what $\ (z^2+\frac 1{z^2})^2\ $ is in relation to $\ z^2+\frac 1{z^2}\ $ and $\ z^4+\frac 1{z^4}\ $.

Now, apply

$$ a^5+a^{-5}\ =\ (a+\frac 1a)\cdot(a^4-a^2+1-a^{-2}+a^{-4}) $$

Good luck!

Answer 4

As $z>0,z+\dfrac1z=4$

Consequently $z,\dfrac1z$ are the roots of $$t^2-4t+1=0$$

$$t^{n+2}=4t^{n+1}-t^n$$

If $z^m+\dfrac1{z^m}=S_m$

$$S_{n+2}=4S_{n+1}-S_n$$ with $S_0=1+1$ and $S_1=4,S_2=14$

Set $n=1,2,3$