Cho

C

- Hướng dẫn giải

Đáp án C.

Ta có ${\left(\frac{z-1}{2z-i}\right)}^4=1$$\iff {\left(z-1\right)}^4-{\left(2z-i\right)}^4=0$. Đặt $f\left(z\right)={\left(z-1\right)}^4-{\left(2z-1\right)}^4$. Phương trình$f\left(z\right)=0$ có 4 nghiệm nên $f\left(z\right)=15\left(z-z_1\right)\left(z-z_2\right)$$\left(z-z_3\right)\left(z-z_4\right)$

Do $i^2=-1$   nên $z^2+1=z^2-i^2$$=\left(z-i\right)\left(z+i\right)$ . Từ đó ta có:

$P=\left[\left(z_1-i\right)\left(z_2-i\right)\left(z_3-i\right)\left(z_4-i\right)\right]$$.\left[\left(z_1+i\right)\left(z_2+i\right)\left(z_3+i\right)\left(z_4+i\right)\right]$

$=\left[\left(i-z_1\right)\left(i-z_2\right)\left(i-z_3\right)\left(i-z_4\right)\right]$$.\left[\left(-i-z_1\right)\left(-i-z_2\right)\left(-i-z_3\right)\left(-i-z_4\right)\right]$

$\Rightarrow P=\frac{f\left(i\right)}{15}.\frac{f\left(-i\right)}{15}$$=\frac{{\left(i-1\right)}^4-{\left(2i-1\right)}^4}{15}.\frac{{\left(-i-1\right)}^4-{\left(-2i-1\right)}^4}{15}$$=\frac{13}{5}$