C)
cách làm:
Đặt z=a+bi;a,b∈R⇒|z−2−3i|=1⇔|(a−2)+(b−3)i|=1z=a+bi;a,b∈R⇒|z−2−3i|=1⇔|(a−2)+(b−3)i|=1
⇔(a−2)2+(b−3)2=1⇔(a−2)2+(b−3)2=1
Đặt a−2=sint;b−3=cost.a−2=sint;b−3=cost. Khi đó ∣∣¯¯¯z+1+i∣∣=|(a+1)+(1−b)i|=√(a+1)2+(1−b)2|z¯+1+i|=|(a+1)+(1−b)i|=(a+1)2+(1−b)2
Ta có (a+1)2+(1−b)2=(sint+3)2+(cost+2)2(a+1)2+(1−b)2=(sint+3)2+(cost+2)2
=14+6sint+4cost≥14+√62+42=14+2√13=14+6sint+4cost≥14+62+42=14+213
Do đó ∣∣¯¯¯z+1+i∣∣≥1+√13|z¯+1+i|≥1+13