A person walks `2sqrt(2)` units away from origin in south west direction `(S45^(@)W)` to reach `A`, then walks `sqrt(2)` units in south east direction `(S45^(@)E)` to reach `B`. From `B` he travel is `4` units horizontally towards east to reach `C`. Then he travels along a circular path with centre at origin through an angle of `2pi//3` in anti-clockwise direction to reach his destination `D`.
Position of `D` in argand plane is (`w` is an imaginary cube root of unity)

Options

A. `(3+i)omega`
B. `-(1+i)omega^(2)`
C. `3(1-i)omega`
D. `(1-3i)omega`

Explanation

To solve the problem step by step, we will determine the coordinates of points A, B, C, and D in the Argand plane.

### Step 1: Determine the coordinates of point A
The person walks `2√2` units in the southwest direction (S45°W). In the Argand plane, this direction corresponds to an angle of 225° (or 5π/4 radians).

Using the polar to rectangular conversion:
- \( A = 2\sqrt{2} \left( \cos(225°) + i \sin(225°) \right) \)

Calculating:
- \( \cos(225°) = -\frac{1}{\sqrt{2}} \)
- \( \sin(225°) = -\frac{1}{\sqrt{2}} \)

Thus,
\[ A = 2\sqrt{2} \left( -\frac{1}{\sqrt{2}} - i \frac{1}{\sqrt{2}} \right) = -2 - 2i \]

### Step 2: Determine the coordinates of point B
From point A, the person walks `√2` units in the southeast direction (S45°E), which corresponds to an angle of 135° (or 3π/4 radians).

Using the polar to rectangular conversion:
- \( B = A + \sqrt{2} \left( \cos(135°) + i \sin(135°) \right) \)

Calculating:
- \( \cos(135°) = -\frac{1}{\sqrt{2}} \)
- \( \sin(135°) = \frac{1}{\sqrt{2}} \)

Thus,
\[ B = (-2 - 2i) + \sqrt{2} \left( -\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \right) = (-2 - 2i) + (-1 + i) = -3 - i \]

### Step 3: Determine the coordinates of point C
From point B, the person travels `4` units horizontally towards the east. Thus, the coordinates of point C will be:
\[ C = B + 4 = (-3 - i) + 4 = 1 - i \]

### Step 4: Determine the coordinates of point D
Now, the person travels along a circular path with center at the origin through an angle of \( \frac{2\pi}{3} \) in an anti-clockwise direction from point C.

The position of point C in polar form is:
- \( r = |C| = \sqrt{1^2 + (-1)^2} = \sqrt{2} \)
- \( \theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4} \)

Now, we add \( \frac{2\pi}{3} \) to the angle:
\[ \theta_D = -\frac{\pi}{4} + \frac{2\pi}{3} = -\frac{3\pi}{12} + \frac{8\pi}{12} = \frac{5\pi}{12} \]

Now, we convert back to rectangular coordinates:
\[ D = r \left( \cos\left(\frac{5\pi}{12}\right) + i \sin\left(\frac{5\pi}{12}\right) \right) = \sqrt{2} \left( \cos\left(\frac{5\pi}{12}\right) + i \sin\left(\frac{5\pi}{12}\right) \right) \]

Calculating:
- \( \cos\left(\frac{5\pi}{12}\right) = \frac{\sqrt{6} - \sqrt{2}}{4} \)
- \( \sin\left(\frac{5\pi}{12}\right) = \frac{\sqrt{6} + \sqrt{2}}{4} \)

Thus,
\[ D = \sqrt{2} \left( \frac{\sqrt{6} - \sqrt{2}}{4} + i \frac{\sqrt{6} + \sqrt{2}}{4} \right) = \frac{\sqrt{12} - \sqrt{4}}{4} + i \frac{\sqrt{12} + \sqrt{4}}{4} = \frac{2\sqrt{3} - 2}{4} + i \frac{2\sqrt{3} + 2}{4} \]
\[ D = \frac{\sqrt{3} - 1}{2} + i \frac{\sqrt{3} + 1}{2} \]

### Final Position of D
Thus, the position of D in the Argand plane is:
\[ D = \frac{\sqrt{3} - 1}{2} + i \frac{\sqrt{3} + 1}{2} \]