If a is a complex number such that `|a|=1`, then find thevalue of a, so that equation `az^2 + z +1=0` has one purely imaginary root.

Explanation

`az^(2) + z +1=0" "(1)`
Taking conjugate on both sides,
`overline(az^(2) + z+1=bar0)`
`rArr bara(barz)^(2) + barz+ bar1 =0`
`baraz^(2) - z+1=0" "("since" barz = -z " is purely imaginary")`
Eliminating z form both the equations, we get
`(bara -a)^(2) + 2(a+bara) = 0`
Let `" "a=cos theta +isin theta" "(because |a| =1)`
`rArr (-2isin theta)^(2) + 2(2cos theta) = 0`
`cos^(2) theta+cos theta -1 =0`
`rArr cos^(2) theta + cos theta -1 =0`
`rArr " " cos theta = (-1pm sqrt(1+4))/(2) =(sqrt(5)-1)/(2)`
Hence, `a =cos theta +isin theta`, where `theta= cos^(-1)((sqrt(5)-1)/(2))`