Given, |z+5|≤4, which is equation of circle. |z+5|≤4 ⇒(x+5)2+y2≤16 and z(1+i)+z(1−i)≥−10 ⇒(z+z)+i(z−z)≥−10 ⇒x−y+5≥0 From Eqs. (i) and (ii), region bounded by inequalities are
Now, |z+1|2=|z−(−1)|2 Maximum value of |z+1|2 will be equal to (AC)2. Now, (x+5)2+y2=16 and x−y+5=0 Given, y=±2√2 and x=±2√2−5 ∴ Coordinates are cA(−2√2−5,−2√2) B(2√2−5,2√2) C(−1,0) Then, AC2=(2√2+4)2+(2√2)2 =32+16√2 Given, that maximum value of |z+1|2 is α+β√2 ⇒α+β√2=32+16√2 ⇒α=32,β=16 ∴α+β=32+16=48