Problem:
Let z=a+bi be the complex number with β£zβ£=5 and b>0 such that the distance between (1+2i)z3 and z5 is maximized, and let z4=c+di. Find c+d.
Solution:
The number z maximizes β£β£β£β(1+2i)z3βz5β£β£β£β=β£zβ£3β β£β£β£β1+2iβz2β£β£β£β. Because β£zβ£3=125 is fixed, it follows that z2 points in the direction of β1β2i with length 25, and thus z2=βΞ΄(1+2i) for some positive real number Ξ΄. Thus 25=Ξ΄β£1+2iβ£=Ξ΄5β and Ξ΄2=125. Finally, z4=125(1+2i)2=125(β3+4i), and the requested sum is β375+500=125β.