If `z_1a n dz_2` both satisfy `z+\bar z =2|z-1|` and `a r g(z_1-z_2)=pi/4` , then find `I m(z_1+z_2)` .

Explanation

Let `z = a+ib,z_{1}=a_{1}+ib_{1}` and` z_{2}=a_{2}+ib_{2}`then

`z+\bar{z}=2\ | z-1 \ |`

`\Rightarrow (a+ib)+(a-ib)=2\| a-1+ib \ |`

`\Rightarrow 2a=1+b^{2}`

So, `z_{1}` and `z_{2}` both satisfy this,

So, we have

`2a_{1}=1+b_{1}^{2}` and `2a_{2}=1+b_{2}^{2}`

`\Rightarrow 2(a_1-a_2)=(b_1+b_2)(b_1-b_2)`

...