Let |z|2+(3−4i)z+(3+4i)¯z−92=0 and (1−i)z+(1+i)¯z−16=0 intersect at z1 and z2. Then the sum of the areas of two parallelograms having origin as one common vertex, and (z1+z2) as diagonal of one parallelogram and (z1−z2) as diagonal of the other parallelogram is

Let |z|2+(3−4i)z+(3+4i)¯z−92=0 and (1−i)z+(1+i)¯z−16=0 inter

1.	Let \|z\|2+(3−4i)z+(3+4i)¯z−92=0 and (1−i)z+(1+i)¯z−16=0 intersect at z1 and z2. Then the sum of the areas of two parallelograms having origin as one common vertex, and (z1+z2) as diagonal of one parallelogram and (z1−z2) as diagonal of the other parallelogram is
Answer» Let $\| z \|^{2} + (3 - 4 i) z + (3 + 4 i) ¯ z - 92 = 0$ and $(1 - i) z + (1 + i) ¯ z - 16 = 0$ intersect at $z_{1}$ and $z_{2} .$ Then the sum of the areas of two parallelograms having origin as one common vertex, and $(z_{1} + z_{2}$ ) as diagonal of one parallelogram and $(z_{1} - z_{2})$ as diagonal of the other parallelogram is