Complex Numbers
Rotation – Compass Bearing Navigation
Complex Numbers_PYQ
Grade 11
Question:
A man walks a distance of $3$ units from the origin towards the North-East (N $45°$ E) direction. From there, he walks a distance of $4$ units towards the North-West (N $45°$ W) direction to reach a point $P$. Then the position of $P$ in the Argand plane is
$3e^{i\pi/4}+4i$
$(3-4i)e^{i\pi/4}$
$(4+3i)e^{i\pi/4}$
$(3+4i)e^{i\pi/4}$
Step-by-Step Solution
Key Concept: Compass bearing movements are multiplications by unit complex numbers. NE=$e^{i\pi/4}$, NW=$e^{i3\pi/4}$. The result $(-1+7i)/\sqrt{2}$ factors neatly as $(3+4i)e^{i\pi/4}$.
**Step 1: Express NE and NW unit vectors**
NE (N45°E) direction: $e^{i\pi/4}=\dfrac{1+i}{\sqrt{2}}$. NW (N45°W) direction: $e^{i\cdot 3\pi/4}=\dfrac{-1+i}{\sqrt{2}}$.
**Step 2: Compute position P**
$P = 3e^{i\pi/4}+4e^{i3\pi/4} = \dfrac{3(1+i)}{\sqrt{2}}+\dfrac{4(-1+i)}{\sqrt{2}} = \dfrac{(3-4)+(3+4)i}{\sqrt{2}} = \dfrac{-1+7i}{\sqrt{2}}$.
**Step 3: Factor as (3+4i)e^{iπ/4}**
$(3+4i)e^{i\pi/4}=(3+4i)\cdot\dfrac{1+i}{\sqrt{2}}=\dfrac{3+3i+4i-4}{\sqrt{2}}=\dfrac{-1+7i}{\sqrt{2}}$. ✓
Correct Answer: 4