Let $z$ be a complex number satisfying $| z | = 2$. In the complex number plane with the origin O, let A and B be the points representing $1 + z$ and $1 - \frac { 1 } { 2 } z$, respectively.
First of all, we can express the complex number $z$ as
$$z = \mathbf { M } ( \cos \theta + i \sin \theta ) \quad ( - \pi \leqq \theta < \pi ) .$$
(1) If $z$ is not a real number, then the area $S$ of the triangle OAB is $S = \mathbf { N }$. For $\mathbf{N}$, choose the correct answer from among (0) $\sim$ (8) below.
Hence, when $\theta = \pm \frac { \mathbf { O } } { \mathbf { P } } \pi$, $S$ is maximized.
(0) $\frac { 1 } { 2 } \left| \sin \left( \theta + \frac { 1 } { 3 } \pi \right) \right|$ (1) $\frac { 1 } { 2 } | \sin \theta |$ (2) $\frac { 1 } { 2 } \left| \sin \left( \theta - \frac { 1 } { 3 } \pi \right) \right|$ (3) $\left| \sin \left( \theta + \frac { 1 } { 3 } \pi \right) \right|$ (4) $| \sin \theta |$ (5) $\left| \sin \left( \theta - \frac { 1 } { 3 } \pi \right) \right|$ (6) $\frac { 3 } { 2 } \left| \sin \left( \theta + \frac { 1 } { 3 } \pi \right) \right|$ (7) $\frac { 3 } { 2 } | \sin \theta |$ (8) $\frac { 3 } { 2 } \left| \sin \left( \theta - \frac { 1 } { 3 } \pi \right) \right|$
(2) When the triangle OAB is an isosceles triangle where $\mathrm { OA } = \mathrm { OB }$, we see that
$$| 1 + z | = \left| 1 - \frac { 1 } { 2 } z \right| = \sqrt { \mathbf { Q } }$$
and
$$\arg ( 1 + z ) = \pm \frac { \mathbf { R } } { \mathbf { S } } \pi , \quad \arg \left( 1 - \frac { 1 } { 2 } z \right) = \mp \frac { \mathbf { T } } { \mathbf{U} } \pi ,$$
where the right-hand sides of the equations are of opposite signs, and where $- \pi \leqq \arg ( 1 + z ) < \pi$ and $- \pi \leqq \arg \left( 1 - \frac { 1 } { 2 } z \right) < \pi$.