Cho hai số phức \(z_1,\,z_2\) thỏa mãn \(\left|z_1\right|=2\), \(\left|z_2\right|=\sqrt{3}\). Gọi \(M,\,N\) là các điểm biểu diễn cho \(z_1\) và \(iz_2\). Biết \(\widehat{MON}=30^\circ\). Tính \(S=\left|z_1^2+4z_2^2\right|\).
 | \(4\sqrt{7}\) |
 | \(3\sqrt{3}\) |
 | \(5\sqrt{2}\) |
 | \(\sqrt{5}\) |
Chọn phương án A.
Giả sử \(z_1=a+bi\), \(z_2=c+di\). Ta có
- \(\left|z_1\right|=\sqrt{a^2+b^2}=2\Leftrightarrow a^2+b^2=4\)
- \(\left|z_2\right|=\sqrt{c^2+d^2}=\sqrt{3}\Leftrightarrow c^2+d^2=3\)
- \(M(a;b)\)
- \(iz_2=ci+di^2=-d+ci\Rightarrow N(-d;c)\)
&\cos\widehat{MON}&=\cos30^\circ\\
\Leftrightarrow&\dfrac{\overrightarrow{OM}\cdot\overrightarrow{ON}}{\left|\overrightarrow{OM}\right|\cdot\left|\overrightarrow{ON}\right|}&=\dfrac{\sqrt{3}}{2}\\
\Leftrightarrow&\dfrac{a\cdot(-d)+b\cdot c}{\sqrt{a^2+b^2}\cdot\sqrt{(-d)^2+c^2}}&=\dfrac{\sqrt{3}}{2}\\
\Leftrightarrow&\dfrac{bc-ad}{\sqrt{a^2+b^2}\cdot\sqrt{c^2+d^2}}&=\dfrac{\sqrt{3}}{2}\\
\Leftrightarrow&\dfrac{bc-ad}{2\sqrt{3}}&=\dfrac{\sqrt{3}}{2}\\
\Leftrightarrow&bc-ad&=3.
\end{eqnarray*}$$
Ta có $$\begin{aligned}
z_1^2+4z_2^2&=z_1^2-\left(2iz_2\right)^2\\
&=\left(z_1-2iz_2\right)\left(z_1+2iz_2\right)\\
&=\left([a+2d]+[b-2c]i\right)\left([a-2d]+[b+2c]i\right)
\end{aligned}$$
\blacksquare\,\left|z_1-2iz_2\right|&=\sqrt{(a+2d)^2+(b-2c)^2}\\
&=\sqrt{\left(a^2+b^2\right)+4\left(c^2+d^2\right)+4(ad-bc)}\\
&=\sqrt{4+4\cdot3+4\left(-3\right)}\\
&=2.
\end{aligned}\)
\(\begin{aligned}
\blacksquare\,\left|z_1+2iz_2\right|&=\sqrt{(a-2d)^2+(b+2c)^2}\\
&=\sqrt{\left(a^2+b^2\right)+4\left(c^2+d^2\right)+4(bc-ad)}\\
&=\sqrt{4+4\cdot3+4\cdot3}\\
&=2\sqrt{7}.
\end{aligned}\)
Khi đó $$\begin{aligned}
S&=\left|z_1^2+4z_2^2\right|\\
&=\left|z_1-2iz_2\right|\cdot\left|z_1+2iz_2\right|\\
&=2\cdot2\sqrt{7}=4\sqrt{7}.
\end{aligned}$$