Let w = z ¯ z + k _ 1 z + k _ 2 iz + λ ( 1 + i ) , k _ 1 , k _ 2 ∈...

Given,

$w = z\overset{¯}{z} + {k}_{1}z + {k}_{2}iz + \lambda ( 1 + i ) , {k}_{1}, {k}_{2} \in \mathbb{R} .$ .

And $Re ( w ) = 0$,

So, taking $z = x + iy \text{ and } \overset{¯}{z} = x - iy$

We get,

$w = {x}^{2} + {y}^{2} + {k}_{1}\left(x + iy\right) + {k}_{2}i\left(x + iy\right) + \lambda ( 1 + i )$

$⇒ Re \left(w\right) = 0 ⇒ {x}^{2} + {y}^{2} + {k}_{1}x - {k}_{2}y + \lambda = 0$

And, $Im ( w ) = 0 ⇒ {k}_{1}y + {k}_{2}x + \lambda = 0$

Now given $Re ( w ) = 0$ circle is touching $y = 1$ and $y -$axis, with radius $1$, so we have equation of circle

${x}^{2} + {y}^{2} - 2x - 4y + 4 = 0$

So, on comparing with $Re ( w ) = 0$ we get,

${k}_{1} = - 2, {k}_{2} = 4, \lambda = 4$

So, $Im ( w ) = 0 ⇒ - 2y + 4x + 4 = 0$

And given $- 2y + 4x + 4 = 0$ intersects the circle,

So, solving ${x}^{2} + {y}^{2} - 2x - 4y + 4 = 0$ and $- 2y + 4x + 4 = 0$ we get,

Intersecting point as $$A\left(0, 2\right)$ and $B\left(\dfrac{2}{5}, \dfrac{14}{5}\right)$

Hence, by distance formula we get,

$30 ( AB{)}^{2} = 30\left[{\left(0 - \dfrac{2}{5}\right)}^{2} + {\left(2 - \dfrac{14}{5}\right)}^{2}\right] = 24