Let z and w be two complex numbers, such that w=zzˉ−2z+2, z−3iz+i=1 and Re(w) has minimum value. Then the minimum value of n∈N for which wn is real, is equal to
Answer: 4
Solutions · 1
Official solution
∣z+i∣=∣z−3i∣ means z lies on the perpendicular bisector of (0,−1) and (0,3), i.e. y=1.
w=∣z∣2−2z+2=(x2+y2−2x+2)+i(−2y). With y=1: w=(x−1)2+2−2i, minimized at x=1, giving w=2−2i=22e−iπ/4.
wn=(22)ne−inπ/4 is real when sin4nπ=0, i.e. n a multiple of 4. Minimum such n∈N is 4.