Given that z2+4|z|2−2zz¯+z¯z−16=0……  (1) Put z=r(cos⁡θ+isin⁡θ)⇒z¯=r(cos⁡θ−isin⁡θ) And zz¯=cos⁡2θ+isin⁡2θ,z¯z=cos⁡2θ−isin⁡2θ|z¯|2=r2 Substitute all these in equation (1) ,we get r2+4r2−4cos⁡2θ−16=0⇒r4+4−r2(4cos⁡2θ+16)=0 Put r2=t⇒t2−t(4cos⁡2θ+16)+4=0⇒t=4cos⁡2θ+16±(4cos⁡2θ+16)2−162⇒r2=2(cos⁡2θ+4)±2(cos⁡2θ+4)2−1∴ Maximum of r2=10+224  , when cos2θ=1=6+4+26×4=(6+4)2∴ Maximum of =2+6