Number of ordered pairs(s) (a, b) of real numbers such that `(a+ib)^2008=a-ib` holds good is

Options

A. 2008
B. 2009
C. 2010
D. 1

Answer

C

Explanation

Let `z=a+ib`
`rArr barz=a -ib`
So, we have
`z^(2008)=barz`
`therefore |z|^(2008)=|barz|=|z|`
`rArr |z|[|z|^(2007)-1]=0`
`rArr |z|=0 or |z|=1`
if `|z|=0, "then "z=0`
If`|z|=1, "then "z^(2009)=zbarz=|z|^(2) =1.`
So, there are 2009 value of z.
Therefore, total number of ordered pairs is 2010.