In argand plane `|z|` represent the distance of a point z from the origin. In general `|z_(1)-z_(2)|` represent the distance between two points `z_(1)` and `z_(2)`. Also for a general moving point z in argand plane, if `arg(z)=theta`, then `z=|z|e^(i theta)`, where `e^(i theta)=cos theta+i sin theta`.
If `z_(1)=4e^(i pi//3)` and `z_(2)=2e^(i5pi//6)`, then `|z_(1)-z_(2)|` equals

Options

A. 20
B. `2sqrt(3)`
C. `sqrt(20)`
D. `20sqrt(3)`

Answer

To find the distance between the points \( z_1 \) and \( z_2 \) in the Argand plane, we will follow these steps:

### Step 1: Write \( z_1 \) and \( z_2 \) in Cartesian form
Given:
- \( z_1 = 4 e^{i \frac{\pi}{3}} \)
- \( z_2 = 2 e^{i \frac{5\pi}{6}} \)

Using the formula \( z = |z| e^{i \theta} = |z| (\cos \theta + i \sin \theta) \):

For \( z_1 \):
\[ z_1 = 4 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right) = 4 \left( \frac{1}{2} + i \frac{\sqrt{3}}{2} \right) = 2 + 2\sqrt{3}i \]

For \( z_2 \):
\[ z_2 = 2 \left( \cos \frac{5\pi}{6} + i \sin \frac{5\pi}{6} \right) = 2 \left( -\frac{\sqrt{3}}{2} + i \frac{1}{2} \right) = -\sqrt{3} + i \]

### Step 2: Calculate \( z_1 - z_2 \)
Now we find \( z_1 - z_2 \):
\[ z_1 - z_2 = (2 + 2\sqrt{3}i) - (-\sqrt{3} + i) = 2 + \sqrt{3} + (2\sqrt{3} - 1)i \]

### Step 3: Find the modulus \( |z_1 - z_2| \)
The distance \( |z_1 - z_2| \) is given by:
\[ |z_1 - z_2| = \sqrt{(2 + \sqrt{3})^2 + (2\sqrt{3} - 1)^2} \]

Calculating each term:
1. \( (2 + \sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3} \)
2. \( (2\sqrt{3} - 1)^2 = 4 \cdot 3 - 4\sqrt{3} + 1 = 12 - 4\sqrt{3} + 1 = 13 - 4\sqrt{3} \)

Now combine these:
\[ |z_1 - z_2|^2 = (7 + 4\sqrt{3}) + (13 - 4\sqrt{3}) = 20 \]
Thus,
\[ |z_1 - z_2| = \sqrt{20} = 2\sqrt{5} \]

### Final Answer
The distance \( |z_1 - z_2| \) equals \( 2\sqrt{5} \).

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