1.

Show that a real value of `x`will satisfy hte equation `(1-i x)//(1+i x)=a-i b`if `a^2+b^2=1,w h e r ea ,b`real.

Answer» We have
`(1-ix)/(1+ix)=(a-ib)/(1)rArr((1-ix)+(1+ix))/((1-ix)-(1+ix))=((a-ib)+1)/((a-ib)-1)" "["by componendo and dividendo"]`
`rArr (2)/(-2ix)=((a+1)-ib)/((a-1)-ib)rArr -ix=((a-1)-ib)/((a+1)-ib)xx((a+1)+ib)/((a+1)+ib)`
`rArr -ix = ((a^(2)-1+b^(2))+{(a-1)b-(a+1)b}i)/((a+1)^(2)+b^(2))`
`=(-2bi)/((a+1)^(2)+b^(2))" "[because a^(2)+b^(2)=1]`
`rArr x=(2b)/((a+1)^(2)+b^(2))`, which is purely real.


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