Let z=x+iy
|z−3−2i|≤2....(Given equation)
⇒|x+iy−3−2i|≤2
⇒|x−3+i(y−2)|≤2
So its a circle with radius 2 where 1≤x≤5 and 0≤y≤4
Now consider |2z−6+5i|≥k
⇒|2(x−3)+i(2y+5)|≥k
⇒√4(x−3)2+(2y+5)2≥k.....(1)
For minimum value of ′k′ either (x−3)2=0 or (2y+5)2=0 or both of them are 0
⇒x−3=0⇒x=3. This is value is possible since 1≤x≤5.
⇒2y+5=0⇒y=−5/2. This value is not possible since 0≤y≤4. So we need to consider minimum value of y that is 0.
Put x=3 and y=0 in equation (1), we get
⇒√0+25=k
⇒k=5