Given:
A=z∈C:z2−(zˉ)2−112iαz−αˉzˉ=1
B=z∈C:∣z+3i∣=4
α=8−14i
Let
z=x+iy
So,
αz=(8−14i)(x+iy)
⇒αz=(8x+14y)+i(−14x+8y)
Also,
z+zˉ=2x and z−zˉ=2iy
Set A:
z2−(zˉ)2−112iαz−αˉzˉ=1
⇒(z+zˉ)(z−zˉ)−112i2i(−14x+8y)=1
⇒4ixy−112i2i(−14x+8y)=1
⇒(2xy−56)−14x+8y=1
⇒xy+7x−4y−28=0
⇒(x−4)(y+7)=0
⇒x=4 or y=−7
Set B:
∣z+3i∣=4
⇒∣x+i(y+3)∣=4
⇒x2+(y+3)2=16
When x=4⇒y=−3
When y=−7⇒x=0
∴A∩B=4−3i,0−7i
So, z∈A∩B∑(Rez−Imz)=4−(−3)+(0−(−7))=14