Cho số phức z thỏa mãn |z^2 -2z+5|=|(z-1+2i)(z+3i-1)|. Tìm giá trị nhỏ nhất của P=|w|.

* Hướng dẫn giải

Đáp án cần chọn là: C

Ta có: $\left|z^2-2z+5\right|=\left|\left(z-1+2i\right)\left(z+3i-1\right)\right|$

$$\begin{gathered} \iff \left|{\left(z-1\right)}^2+4\right|=\left|\left(z-1+2i\right)\left(z+3i-1\right)\right| \\ \iff \left|{\left(z-1\right)}^2-{\left(2i\right)}^2\right|=\left|\left(z-1+2i\right)\left(z+3i-1\right)\right| \\ \iff \left|\left(z-1+2i\right)\left(z-1-2i\right)\right|=\left|\left(z-1+2i\right)\left(z+3i-1\right)\right| \\ \iff \begin{array}{rcl}z-1+2i& =& 0\left(1\right)\\ \left|\left(z-1-2i\right)\right|& =& \left|\left(z+3i-1\right)\right|\left(2\right)\end{array} \end{gathered}$$

Từ (1) $\Rightarrow z=1-2i\Rightarrow w=-1\Rightarrow P=\left|w\right|=1$.

Xét (2). Gọi $z=x+yi(x,y\in R)$

Ta có:

$\left|\left(z-1-2i\right)\right|=\left|\left(z+3i-1\right)\right|\iff {\left(x-1\right)}^2+{\left(y-2\right)}^2={\left(x-1\right)}^2+{\left(y+3\right)}^2\iff y=\frac{-1}{2}$

Khi đó $w=x-\frac{1}{2}i-2+2i=\left(x-2\right)+\frac{3}{2}i\Rightarrow P=\left|w\right|=\sqrt{{\left(x-2\right)}^2+{\left(\frac{3}{2}\right)}^2}\ge \frac{3}{2}>1$

Vậy $P_{min}=1$.