Cho hai số phức

A

- Hướng dẫn giải

Đáp án A.

Từ  

$\begin{array}{l}z=\frac{z_1}{z_2}=a+bi\\ \to \left|z\right|=\left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|}=\sqrt{a^2+b^2}\\ \to \sqrt{a^2+b^2}=\frac{3}{4}\end{array}$

Từ  

$\begin{array}{l}\frac{\left|z_1-z_2\right|}{\left|z_2\right|}=\left|\frac{z_1-z_2}{z_2}\right|\\ =\left|\frac{z_1}{z_2}-1\right|\\ =\left|z-1\right|=\frac{\sqrt{37}}{4}\\ \to \sqrt{{\left(a-1\right)}^2+b^2}=\frac{\sqrt{37}}{4}\end{array}$

Ta có hệ phương trình sau

$\begin{array}{l}\left\{\begin{array}{l}{a}^2+b^2=\frac{9}{16}\\ {\left(a-1\right)}^2+b^2=\frac{37}{16}\end{array}\right.\\ \iff \left\{\begin{array}{l}{a}^2+b^2=\frac{9}{16}\\ {\left(a-1\right)}^2-a^2=\frac{7}{4}\end{array}\right.\\ \iff \left\{\begin{array}{l}{a}^2+b^2=\frac{9}{16}\\ -2a=\frac{3}{4}\end{array}\right.\end{array}$

 $\iff \left\{\begin{array}{l}a=-\frac{3}{8}\\ b^2=\frac{9}{16}-{\left(-\frac{3}{8}\right)}^2=\frac{27}{64}\end{array}\right.$ 

Vậy $b=\pm \frac{3\sqrt{3}}{8}\to \left|b\right|=\frac{3\sqrt{3}}{8}$