Google
Câu hỏi: Hai số phức $z$, $\text{w}$ thay đổi nhưng luôn thỏa mãn đẳng thức $\left( 1+i \right)\left| {{z}^{2}}-2iz-1 \right|=\dfrac{\left| 2022.\overline{z}+2022 \right|}{\text{w}}+2-2i$. Giá trị lớn nhất của $\left| \text{w} \right|$ là
A. $\dfrac{2021\sqrt{2}}{4}$.
B. $\dfrac{1011\sqrt{2}}{2}$.
C. $\dfrac{2023\sqrt{2}}{4}$.
D. $2019$.
A. $\dfrac{2021\sqrt{2}}{4}$.
B. $\dfrac{1011\sqrt{2}}{2}$.
C. $\dfrac{2023\sqrt{2}}{4}$.
D. $2019$.
Ta có: $\left| z-i \right|=\left| \overline{z}+i \right|$ nên $\left| {{z}^{2}}-2iz-1 \right|={{\left| z-i \right|}^{2}}={{\left| \overline{z}+i \right|}^{2}}$.
Phương trình $\left( 1+i \right)\left| {{z}^{2}}-2iz-1 \right|=\dfrac{\left| 2022.\overline{z}+2022 \right|}{\text{w}}+2-2i$
$\Leftrightarrow \left( 1+i \right){{\left| \overline{z}+i \right|}^{2}}=\dfrac{\left| 2022\left( \overline{z}+1 \right) \right|}{\text{w}}+2-2i$ $\Leftrightarrow \left( {{\left| \overline{z}+i \right|}^{2}}-2 \right)+\left( {{\left| \overline{z}+i \right|}^{2}}+2 \right)i=\dfrac{\left| 2022\left( \overline{z}+i \right) \right|}{\text{w}}$ $\left( 1 \right)$.
Điều kiện: $\text{w}\ne 0$ suy ra $\overline{z}+i\ne 0$ hay $\left| \overline{z}+i \right|>0$.
Đặt $t=\left| \overline{z}+i \right|$, $t>0$ ta có phương trình $\left( 1 \right)$ $\Leftrightarrow \left( {{t}^{2}}-2 \right)+\left( {{t}^{2}}+2 \right)i=\dfrac{\left| 2022\left( \overline{z}+i \right) \right|}{\text{w}}$
$\Rightarrow \sqrt{{{\left( {{t}^{2}}-2 \right)}^{2}}+{{\left( {{t}^{2}}+2 \right)}^{2}}}=\dfrac{2022t}{\left| \text{w} \right|}\Leftrightarrow \left| \text{w} \right|=2022\sqrt{\dfrac{{{t}^{2}}}{2\left( {{t}^{4}}+4 \right)}}=1011\sqrt{2}\sqrt{\dfrac{1}{{{t}^{2}}+\dfrac{4}{{{t}^{2}}}}}$ $\Leftrightarrow \left| \text{w} \right|\le 1011\sqrt{2}.\sqrt{\dfrac{1}{2\sqrt{{{t}^{2}}.\dfrac{4}{{{t}^{2}}}}}}=\dfrac{1011\sqrt{2}}{2}$ dấu bằng xảy ra khi ${{t}^{2}}=\dfrac{4}{{{t}^{2}}}$ $\Leftrightarrow \left| \overline{z}+i \right|=\sqrt[2]{2}$ $\Leftrightarrow \text{w}=-\dfrac{1011\sqrt{2}}{2}i$.
Phương trình $\left( 1+i \right)\left| {{z}^{2}}-2iz-1 \right|=\dfrac{\left| 2022.\overline{z}+2022 \right|}{\text{w}}+2-2i$
$\Leftrightarrow \left( 1+i \right){{\left| \overline{z}+i \right|}^{2}}=\dfrac{\left| 2022\left( \overline{z}+1 \right) \right|}{\text{w}}+2-2i$ $\Leftrightarrow \left( {{\left| \overline{z}+i \right|}^{2}}-2 \right)+\left( {{\left| \overline{z}+i \right|}^{2}}+2 \right)i=\dfrac{\left| 2022\left( \overline{z}+i \right) \right|}{\text{w}}$ $\left( 1 \right)$.
Điều kiện: $\text{w}\ne 0$ suy ra $\overline{z}+i\ne 0$ hay $\left| \overline{z}+i \right|>0$.
Đặt $t=\left| \overline{z}+i \right|$, $t>0$ ta có phương trình $\left( 1 \right)$ $\Leftrightarrow \left( {{t}^{2}}-2 \right)+\left( {{t}^{2}}+2 \right)i=\dfrac{\left| 2022\left( \overline{z}+i \right) \right|}{\text{w}}$
$\Rightarrow \sqrt{{{\left( {{t}^{2}}-2 \right)}^{2}}+{{\left( {{t}^{2}}+2 \right)}^{2}}}=\dfrac{2022t}{\left| \text{w} \right|}\Leftrightarrow \left| \text{w} \right|=2022\sqrt{\dfrac{{{t}^{2}}}{2\left( {{t}^{4}}+4 \right)}}=1011\sqrt{2}\sqrt{\dfrac{1}{{{t}^{2}}+\dfrac{4}{{{t}^{2}}}}}$ $\Leftrightarrow \left| \text{w} \right|\le 1011\sqrt{2}.\sqrt{\dfrac{1}{2\sqrt{{{t}^{2}}.\dfrac{4}{{{t}^{2}}}}}}=\dfrac{1011\sqrt{2}}{2}$ dấu bằng xảy ra khi ${{t}^{2}}=\dfrac{4}{{{t}^{2}}}$ $\Leftrightarrow \left| \overline{z}+i \right|=\sqrt[2]{2}$ $\Leftrightarrow \text{w}=-\dfrac{1011\sqrt{2}}{2}i$.
Đáp án B.
