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The value of ( {1-i}{1+i} )^{10}+ ( {1+i}{1-i} )^8=

Question

The value of $(\frac{1 - i}{1 + i})^{10} + (\frac{1 + i}{1 - i})^{8} =$

0

2 i

i

- i

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The value of

(\frac{1 - i}{1 + i})^{10} + (\frac{1 + i}{1 - i})^{8} =

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Text solutionVerified

\frac{1 - i}{1 + i} = \frac{( 1 - i ) ^{2}}{1 ^{2} - i ^{2}} = \frac{1 - 1 - 2 i}{2} = - i

\frac{1 + i}{1 - i} = \frac{( 1 + i ) ^{2}}{1 ^{2} - i ^{2}} = \frac{1 - 1 + 2 i}{2} = i

∴ (\frac{1 - i}{1 + i})^{10} + (\frac{1 + i}{1 - i})^{8}

= (- i)^{10} + (i)^{8}

= i^{10} + i^{8}

= - 1 + 1

- (

i^{4 n - 2} = - 1

i^{4 n} = 8

)

= 0

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