Points `z` in the complex plane satisfying ` "Re"(z+1)^(2)=|z|^(2)+1` lie on

Answer

`a to p; b to p; c to q; d to s`

Explanation

a. `x = sin theta, y = cos theta`
`P = ( 3 sin theta - 4 sin ^(3)theta )^(2) + (3 cos theta - 4 cos^(3) theta )^(2)`
` = sin ^(2) 3 theta + cos ^(2) 3 theta =1 `
b. On adding, we get `a = ( 3- cos 4 theta + 4sin 2theta)/(2) = (1+ sin 2theta )^(2) `
On subtracting, we get `b = (1- sin 2 theta)^(2)`
`rArr ab = cos ^(4) 2theta le 1`
c. `3 cos theta = x^(2) - 8x + 19`
`rArr 3 cos theta = (x - 4)^(2)+ 3 `
Now L.H.S. = `3 cos theta le 3`
or L.H.S. has the greatest value 3.
But R.H.S. = ` (x -4)^(2) + 3 ge 3`
or R.H.S. has the least value 3.
Hence, L.H.S. = R.H.S.
when `3cos theta = (x -4)^(2) + 3=3`
`rArr cos theta = 1 and x -4=0`
`rArr = 2npi and x =4, ` where `n in Z`.
d. Let `lamda = tan theta `
`therfore x = 2 sin 2theta and y = 2cos 2 theta `
`E = x^(2) - xy + y^(2) = 4-4 sin 2 theta cos 2 theta = 4 - 2 sin 4theta`
`E in [2, 6] rArr a + b =8`