Gọi $z = x + yi$ với $x,y \in R$.
Ta có:
$\left| z \right| = \sqrt {{x^2} +
{y^2}} $
${z^2} + 4 = {x^2} - {y^2} + 4 + 2xyi$
$\left| {{z^2} + 4} \right| = 2\left| z
\right| \Leftrightarrow {({x^2} - {y^2} + 4)^2} + 4{x^2}{y^2} = 4({x^2} +
{y^2})$
$\Leftrightarrow {x^4} + {y^4} + 16 +
8{x^2} - 8{y^2} - 2{x^2}{y^2} + 4{x^2}{y^2} - 4{x^2} - 4{y^2} = 0$
$\Leftrightarrow {x^4} + {y^4} +
2{x^2}{y^2} + 4{x^2} - 12{y^2} + 16 = 0$
$\Leftrightarrow {({x^2} + {y^2})^2} +
4{x^2} - 12({\left| z \right|^2} - {x^2}) + 16 = 0$
$\Leftrightarrow {\left| z \right|^4} -
12{\left| z \right|^2} + 16{x^2} + 16 = 0$
$\Delta ' = 36 - (16{x^2} + 16) = 20 -
16{x^2}$
Do ${\left| z \right|^2}$ tồn tại nên $20 - 16{x^2} \ge 0 \Leftrightarrow \frac{{ -
\sqrt 5 }}{2} \le x \le \frac{{\sqrt 5 }}{2}$.
Khi đó: ${\left| z \right|^2} = 6 \pm
\sqrt {20 - 16{x^2}} $
$\Leftrightarrow \left| z \right| =
\sqrt {6 \pm \sqrt {20 - 16{x^2}} } $
$Min\left| z \right| = \sqrt {6 - \sqrt
{20} } = \sqrt 5 - 1 \Leftrightarrow x = 0,y = \sqrt 5 - 1 \Leftrightarrow z = (\sqrt 5 - 1)i$
$Max\left| z \right| = \sqrt {6 + \sqrt
{20} } = \sqrt 5 + 1 \Leftrightarrow x = 0,y = \sqrt 5 + 1 \Leftrightarrow z = (\sqrt 5 + 1)i$
times="" new="" roman""="">Gọi $z = x + yi$ với $x,y \in R$.
times="" new="" roman""="">Ta có:
times="" new="" roman""="">$\left| z \right| = \sqrt {{x^2} +
{y^2}} $
times="" new="" roman""="">${z^2} + 4 = {x^2} - {y^2} + 4 + 2xyi$
times="" new="" roman""="">$\left| {{z^2} + 4} \right| = 2\left| z
\right| \Leftrightarrow {({x^2} - {y^2} + 4)^2} + 4{x^2}{y^2} = 4({x^2} +
{y^2})$
times="" new="" roman""="">$\Leftrightarrow {x^4} + {y^4} + 16 +
8{x^2} - 8{y^2} - 2{x^2}{y^2} + 4{x^2}{y^2} - 4{x^2} - 4{y^2} = 0$
times="" new="" roman""="">$\Leftrightarrow {x^4} + {y^4} +
2{x^2}{y^2} + 4{x^2} - 12{y^2} + 16 = 0$
times="" new="" roman""="">$\Leftrightarrow {({x^2} + {y^2})^2} +
4{x^2} - 12({\left| z \right|^2} - {x^2}) + 16 = 0$
times="" new="" roman""="">$\Leftrightarrow {\left| z \right|^4} -
12{\left| z \right|^2} + 16{x^2} + 16 = 0$
times="" new="" roman""="">$\Delta ' = 36 - (16{x^2} + 16) = 20 -
16{x^2}$
times="" new="" roman""="">Do ${\left| z \right|^2}$ tồn tại nên $20 - 16{x^2} \ge 0 \Leftrightarrow \frac{{ -
\sqrt 5 }}{2} \le x \le \frac{{\sqrt 5 }}{2}$.
times="" new="" roman""="">Khi đó: ${\left| z \right|^2} = 6 \pm
\sqrt {20 - 16{x^2}} $
times="" new="" roman""="">$\Leftrightarrow \left| z \right| =
\sqrt {6 \pm \sqrt {20 - 16{x^2}} } $ trong đó $\frac{{ - \sqrt 5 }}{2} \le x \le \frac{{\sqrt 5 }}{2}$
times="" new="" roman""="">$Min\left| z \right| = \sqrt {6 - \sqrt
{20} } = \sqrt 5 - 1 \Leftrightarrow x = 0,y = \sqrt 5 - 1 \Leftrightarrow z = (\sqrt 5 - 1)i$
times="" new="" roman""="">$Max\left| z \right| = \sqrt {6 + \sqrt
{20} } = \sqrt 5 + 1 \Leftrightarrow x = 0,y = \sqrt 5 + 1 \Leftrightarrow z = (\sqrt 5 + 1)i$