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

Jun 26, 2020

#1
+25490
+3

If

$$z$$ is a complex number satisfying $$z + \dfrac{1}{z} =1$$, calculate $$z^{10}+ \dfrac{1}{z^{10}}$$

$$\begin{array}{|rcll|} \hline \left( z+\dfrac{1}{z} \right)^2 &=& z^2+\dfrac{1}{z^2} + \dfrac{2z}{z} \\\\ 1^2 &=& z^2+\dfrac{1}{z^2} + 2 \\\\ z^2+\dfrac{1}{z^2} &=& 1-2 \\\\ \mathbf{z^2+\dfrac{1}{z^2}} &=& \mathbf{-1} \\ \hline \end{array} \begin{array}{|rcll|} \hline \left( z^2+\dfrac{1}{z^2} \right)\left( z+\dfrac{1}{z} \right) &=& z^3+\dfrac{1}{z^3} + z+\dfrac{1}{z} \\\\ (-1)*1 &=& z^3+\dfrac{1}{z^3} + 1 \\\\ z^3+\dfrac{1}{z^3} &=&-1-1 \\\\ \mathbf{z^3+\dfrac{1}{z^3}} &=& \mathbf{-2} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline \left( z^2+\dfrac{1}{z^2} \right)\left( z^3+\dfrac{1}{z^3} \right) &=& z^5+\dfrac{1}{z^5} + z+\dfrac{1}{z} \\\\ (-1)*(-2) &=& z^5+\dfrac{1}{z^5} + 1 \\\\ z^5+\dfrac{1}{z^5} &=&2-1 \\\\ \mathbf{z^5+\dfrac{1}{z^5}} &=& \mathbf{1} \\ \hline \end{array} \begin{array}{|rcll|} \hline \left( z^5+\dfrac{1}{z^5} \right)^2 &=& z^{10}+\dfrac{1}{z^{10}} +\dfrac{2z^5}{z^5} \\\\ 1^2 &=& z^{10}+\dfrac{1}{z^{10}} + 2 \\\\ z^{10}+\dfrac{1}{z^{10}} &=&1-2 \\\\ \mathbf{z^{10}+\dfrac{1}{z^{10}}} &=& \mathbf{-1} \\ \hline \end{array}$$

edited: Thank you Alan, thank you MaxWong

Jun 26, 2020
edited by heureka  Jun 26, 2020
edited by heureka  Jun 29, 2020
edited by heureka  Jun 29, 2020
edited by heureka  Jun 29, 2020
#3
+8336
+5

heureka:

$$\left(z^2 + \dfrac1{z^2}\right)\left(z^5 + \dfrac1{z^5}\right) = z^{\color{red}7\color{black}} + \dfrac1{z^{\color{red}7\color{black}}} + z^3 + \dfrac1{z^3}$$

MaxWong  Jun 26, 2020
#5
+25490
+4

Thank you, MaxWong

heureka  Jun 29, 2020
#2
+8336
+3

An interesting theorem: If $$z + \dfrac1z = 2\cos \alpha$$, then $$z^n + \dfrac1{z^n} = 2\cos n\alpha$$ for positive integers n and real $$\alpha$$.

(If you don't like to read proofs, then you can jump to the point where I marked "END OF PROOF".)

Proof of the theorem:

We use strong induction.

Base case #1 : n = 1.

Substituting n = 1 and checking shows that base case #1 is true.

Base case #2 : n = 2.

$$z^2 + \dfrac1{z^2} = \left(z + \dfrac1z\right)^2 - 2 = (2\cos \alpha)^2 - 2=2(2\cos^2 \alpha - 1) = 2\cos 2\alpha$$

So base case #2 is true.

Inductive step: Assume $$z^t + \dfrac1{z^t} = 2\cos t\alpha$$ and $$z^{t - 1} + \dfrac1{z^{t - 1}} = 2\cos(t -1)\alpha$$ for some positive integer $$t$$.

$$\quad z^{t + 1} + \dfrac1{z^{t + 1}}\\ = \left(z^t + \dfrac1{z^t}\right)\left(z + \dfrac1z\right) - \left(z^{t - 1} + \dfrac1{z^{t - 1}}\right)\\ = 2\cos t\alpha \cdot 2\cos \alpha - 2\cos (t - 1)\alpha\\ \text{By product to sum formula:}\\ = 2 \left(\cos (t + 1)\alpha + \cos(t - 1)\alpha\right) - 2\cos(t-1)\alpha\\ =2\cos(t + 1)\alpha$$

Therefore by principles of mathematical induction, the proposition is true.

-------------------------------------------------------------------------(END OF PROOF)-------------------------------------------------------------------------

Back to the problem. Using the theorem I mentioned above, and substituting $$\alpha = 60^\circ$$ into the formula,

$$\quad z^{10} + \dfrac1{z^{10}}\\ = 2\cos\left(10\cdot 60^\circ\right)\\ = 2 \cos\left(600^\circ\right)\\ = 2 \cos 240^\circ\\ = \boxed{-1}$$

Jun 26, 2020
edited by MaxWong  Jun 26, 2020
#4
+43
+7

Thank you so much!

Jun 27, 2020