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a=sqrt(3)+1, b=sqrt(3)-1, ((a+bi)/(a-bi)+(b+ai)/(b-ai))^10

Guest Aug 8, 2015

Best Answer 

 #1
avatar+93644 
+13

a=sqrt(3)+1, b=sqrt(3)-1, ((a+bi)/(a-bi)+(b+ai)/(b-ai))^10

 

$$\\\left(\frac{(a+bi)}{(a-bi)}+\frac{(b+ai)}{(b-ai)}\right)^{10} \\\\
=\left(\frac{(a+bi)(a+bi)}{(a-bi)(a+bi)}+\frac{(b+ai)(b+ai)}{(b-ai)(b+ai)}\right)^{10} \\\\
=\left(\frac{a^2-b^2+2abi}{a^2+b^2}+\frac{b^2-a^2+2abi}{b^2+a^2}\right)^{10} \\\\
=\left(\frac{a^2-b^2+2abi+b^2-a^2+2abi}{a^2+b^2}\right)^{10} \\\\
=\left(\frac{4abi}{a^2+b^2}\right)^{10} \\\\
$We are given that $\;\;\;\ a=(\sqrt3+1)\qquadand\qquad b=(\sqrt3-1)\\\\
=\left(\frac{4(\sqrt3+1)(\sqrt3-1)i}{(\sqrt3+1)^2+(\sqrt3-1)^2}\right)^{10} \\\\$$

$$\\=\left(\frac{4(3-1)i}{(3+1+2\sqrt3)+(3+1-2\sqrt3)}\right)^{10} \\\\
=\left(\frac{4(2)i}{8}\right)^{10} \\\\
=i^{10} \\\\
=(i^2)^5 \\\\
=(-1)^5 \\\\
=-1$$

Melody  Aug 8, 2015
 #1
avatar+93644 
+13
Best Answer

a=sqrt(3)+1, b=sqrt(3)-1, ((a+bi)/(a-bi)+(b+ai)/(b-ai))^10

 

$$\\\left(\frac{(a+bi)}{(a-bi)}+\frac{(b+ai)}{(b-ai)}\right)^{10} \\\\
=\left(\frac{(a+bi)(a+bi)}{(a-bi)(a+bi)}+\frac{(b+ai)(b+ai)}{(b-ai)(b+ai)}\right)^{10} \\\\
=\left(\frac{a^2-b^2+2abi}{a^2+b^2}+\frac{b^2-a^2+2abi}{b^2+a^2}\right)^{10} \\\\
=\left(\frac{a^2-b^2+2abi+b^2-a^2+2abi}{a^2+b^2}\right)^{10} \\\\
=\left(\frac{4abi}{a^2+b^2}\right)^{10} \\\\
$We are given that $\;\;\;\ a=(\sqrt3+1)\qquadand\qquad b=(\sqrt3-1)\\\\
=\left(\frac{4(\sqrt3+1)(\sqrt3-1)i}{(\sqrt3+1)^2+(\sqrt3-1)^2}\right)^{10} \\\\$$

$$\\=\left(\frac{4(3-1)i}{(3+1+2\sqrt3)+(3+1-2\sqrt3)}\right)^{10} \\\\
=\left(\frac{4(2)i}{8}\right)^{10} \\\\
=i^{10} \\\\
=(i^2)^5 \\\\
=(-1)^5 \\\\
=-1$$

Melody  Aug 8, 2015

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