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# Algebra

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If z is a complex number satisfying z + 1/z =1, calculate $$z^{12} + 1/z^{12}$$

Jan 15, 2022

#1
+118117
+1

There might be an easier way and you need to check my working for careless errors

$$z+\frac{1}{z}=1\\ \left(z+\frac{1}{z}\right)^2=1^2\\ \left(z^2+\frac{1}{z^2}+2\right)=1\\ \left(z^2+\frac{1}{z^2}\right)=-1\\ \left(z^2+\frac{1}{z^2}\right)\left(z+\frac{1}{z}\right)=-1\left(z+\frac{1}{z}\right)\\ \left(z^3+z+\frac{1}{z}+\frac{1}{z^3}\right)=-1*1\\ \left(z^3+\frac{1}{z^3}+1\right)=-1\\ \left(z^3+\frac{1}{z^3}\right)=-2\\ \left(z^3+\frac{1}{z^3}\right)^2=(-2)^2\\ \left(z^6+\frac{1}{z^6}+2\right)=4\\ \left(z^6+\frac{1}{z^6}\right)=2\\ \left(z^6+\frac{1}{z^6}\right)^2=2^2\\ \left(z^{12}+\frac{1}{z^{12}}+2\right)=4\\~\\ z^{12}+\frac{1}{z^{12}}=2\\$$

LaTex:

z+\frac{1}{z}=1\\
\left(z+\frac{1}{z}\right)^2=1^2\\
\left(z^2+\frac{1}{z^2}+2\right)=1\\
\left(z^2+\frac{1}{z^2}\right)=-1\\
\left(z^2+\frac{1}{z^2}\right)\left(z+\frac{1}{z}\right)=-1\left(z+\frac{1}{z}\right)\\
\left(z^3+z+\frac{1}{z}+\frac{1}{z^3}\right)=-1*1\\
\left(z^3+\frac{1}{z^3}+1\right)=-1\\
\left(z^3+\frac{1}{z^3}\right)=-2\\
\left(z^3+\frac{1}{z^3}\right)^2=(-2)^2\\
\left(z^6+\frac{1}{z^6}+2\right)=4\\
\left(z^6+\frac{1}{z^6}\right)=2\\
\left(z^6+\frac{1}{z^6}\right)^2=2^2\\
\left(z^{12}+\frac{1}{z^{12}}+2\right)=4\\~\\
z^{12}+\frac{1}{z^{12}}=2\\

Jan 15, 2022