Answer:
Step-by-step explanation:
[tex]z + \dfrac{1}{z}=3\\\\\\Square \ both \ sides\\\\(z+\dfrac{1}{z})^{2}=3^{2}\\\\\\z^{2}+\dfrac{1}{z^{2}}+2*z*\dfrac{1}{z}=9\\\\\\z^{2}+\dfrac{1}{z^{2}}+2=9\\\\\\z^{2}+\dfrac{1}{z^{2}}=9-2\\\\\\z^{2}+\dfrac{1}{z^{2}}=7\\\\\\Now, \ again \ square \ both \ sides,\\\\\\(z^{2}+\dfrac{1}{z^{2}})^{2}=7^{2}\\\\\\(z)^{4}+\dfrac{1}{z^{4}}+2*z^{2}*\dfrac{1}{z^{2}}=49\\\\\\z^{4}+\dfrac{1}{z^{4}}+2=49\\\\\\z^{4}+\dfrac{1}{z^{4}}=49-2\\\\z^{4}+\dfrac{1}{z^{4}}=47[/tex]
