For a∈ mathbbC, let A=z∈ mathbbC:Re(a+ z)>Im( a+z) and B=z∈ mathbbC:Re(a+ z)<Im( a+z). Then among the two statements: (S1) : If Re(a),Im(a)>0, then…

Let $a={x}_{1}+i{y}_{1},\bar{a}={x}_{1}-i{y}_{1}$a=x1​+iy1​,aˉ=x1​−iy1​

And $z={x}_{2}+i{y}_{2},\bar{z}={x}_{2}-i{y}_{2}$z=x2​+iy2​,zˉ=x2​−iy2​

Now,

$$a+\bar{z}=({x}_{1}+{x}_{2})+i({y}_{1}-{y}_{2})$$a+zˉ=(x1​+x2​)+i(y1​−y2​)

and $\bar{a}+z=({x}_{1}+{x}_{2})+i({y}_{2}-{y}_{1})$aˉ+z=(x1​+x2​)+i(y2​−y1​)

Now according to the question,

$Re(a+\bar{z})={x}_{1}+{x}_{2}$Re(a+zˉ)=x1​+x2​ and

$$Im(\bar{a}+z)={y}_{2}-{y}_{1}$$Im(aˉ+z)=y2​−y1​

$$A={z:{x}_{1}+{x}_{2}>{y}_{2}-{y}_{1}}={z:{x}_{1}+{y}_{1}+{x}_{2}>{y}_{2}}$$A=z:x1​+x2​>y2​−y1​=z:x1​+y1​+x2​>y2​

$$B={z:{x}_{1}+{x}_{2}<{y}_{2}-{y}_{1}}={z:{x}_{1}+{y}_{1}+{x}_{2}<{y}_{2}}$$B=z:x1​+x2​<y2​−y1​=z:x1​+y1​+x2​<y2​

If ${y}_{2}=0$y2​=0 and ${x}_{1},{y}_{1}>0$x1​,y1​>0 then

$$A={z:{x}_{2}>-({x}_{1}+{y}_{1})}$$A=z:x2​>−(x1​+y1​)

A covers a part of negative real axis and therefore, does not contain whole real axis

Similarly if ${y}_{2}=0$y2​=0, and ${x}_{1},{y}_{1}<0$x1​,y1​<0, then

$$B={z:{x}_{2}<-({x}_{1}+{y}_{1})}$$B=z:x2​<−(x1​+y1​)

$\therefore$∴ $B$B covers part of positive real axis and therefore does not cover whole real axis.

Hence, both are false.