A particle P starts from $Z_0 = 1 + 2i$. It moves horizontally away from origin by 5 units, then vertically up by 3 units to $Z_1$. From $Z_1$ it moves $\sqrt{2}$ units in direction $\hat{i} + \hat{j}$, then moves through $\pi/2$ anticlockwise on a circle with centre at origin to reach $Z_2$. Then $Z_2 = \dots$
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Argand plane geometric transformations are incredibly simple: Translation by $(a,b)$ is just adding $(a+bi)$. Rotation around the origin by an angle $\theta$ is just multiplying by $e^{i\theta} = (\cos\theta + i\sin\theta)$. For a pure $90^\circ$ turn, just multiply by $i$.
Step 1: Understanding the Concept:
Horizontal/vertical shifts add directly to coordinates. Rotation by $\pi/2$ anticlockwise about origin corresponds to multiplying by $i$. Step 2: Formula Application:
$Z_0 = 1 + 2i$.
Shift 5 units right: $1+5 = 6$. Shift 3 units up: $2+3 = 5$.
$Z_1 = 6 + 5i$.
Vector $\hat{i} + \hat{j}$ has magnitude $\sqrt{2}$. Moving $\sqrt{2}$ in that direction is just adding $1+i$.
New point $= (6+1) + (5+1)i = 7 + 6i$. Step 3: Explanation:
Rotation of $Z = x + iy$ by $\pi/2$ anticlockwise results in $Z_{new} = Z \cdot e^{i\pi/2} = Z \cdot i$.
$Z_2 = (7 + 6i) \cdot i = 7i + 6i^2 = -6 + 7i$. (Re-calculating: $x$ becomes $-y$, $y$ becomes $x$).
If $Z = 7 + 6i$, then $Z_2 = -6 + 7i$. Step 4: Final Answer:
The point $Z_2$ is $-6 + 7i$ (Option C).