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What is the value of the expression `ilog(x-i)` `+ i^2pi` `+ i^3log(x+i)`...
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We want the value of the expression
`ilog(x-i) + i^2pi + i^3log(x+i) + i^4(2tan^(-1)x)`
given that `x>0`.
Firstly consider the complex inverse tangent function
`tan^(-1)z = i/2[log(1-iz) -log(1+iz)]`
By simple algebraic manipulation (using the fact that `i^2=-1` and laws of logarithms) we can rewrite this as
`tan^(-1)z = i/2[log((z+i)/i) - log((z-i)/(-i))]`
`= i/2[log(z+i) - log(z-i) - log(i) + log(-i)]`
`2tan^(-1)x = ilog(x+i) - ilog(x-i) - ilog(i) + ilog(-i)`
Noting that `i^3 = i(i^2) = -i` and further that `i^4 = i(i^3) = -i^2 = 1` we see that ` ` the term `i^(4)2tan^(-1)x` in the original expression cancels out the ` ` logarithmic terms in that expression.
We are left with finding the value of
`i^2pi - ilog(i) + ilog(-i)`
Now, writing `i` and `-i` in polar coordinates on the complex plane, as
`i = e^(i(pi/2))`
`-i = e^(-i(pi/2))`
gives us that `log(i) = i(pi/2)` and `log(-i) = -i(pi/2)` .
Therefore, `-ilog(i) +ilog(-i) = -2i^2(pi/2) = 2(pi/2) = pi`
Finally, the expression
`i^2pi - ilog(i) + ilog(-i)` equates to (`-pi + pi `), ie zero.
The answer is A) 0
Posted by mathsworkmusic on August 21, 2013 at 2:39 PM (Answer #1)
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