If (4+i)(z+z¯)−(3+i)(z−z¯)+26i=0, then the value of z2 is
13
17
19
11
Let z=x+iy, then
2(4+i)x−(3+i)2iy+26i=0⇒4x+y=0,x−3y+13=0⇒x=−1,y=4∴|z|2=17