Question

The real part of the principle value of ` 2^( - i)` is
A. ` sin ( log 2 ) `
B. ` cos ((1 )/( log 2 )) `
C. ` cos [ log ((1 )/( 2 )) ] `
D. ` cos ( log 2 ) `

Answer 1

Correct Answer - C
Let ` z = 2^( - i)`
Taking log on both sides , we get
`log z = log ( 2 ^( - i)) `
` rArr log z = - ilog 2 `
` rArr log z = ilog ((1 )/( 2 ) ) `
` rArr z = e^(ilog (1//2)) ( because e ^(itheta ) = cos theta + isin theta ) `
` rArr z = cos ( log (1 ) / (2) ) + isin (log (1 )/ ( 2 )) `
` therefore ` The real part of ` z = 2^(-i) ` is ` cos (log (1 )/( 2 ))`