Question

Let \(z_1\) and \(z_2\) be two complex numbers, then dot product of \(z_1\) and \(z_2\) is :

• A. \(|z_1||z_2|\)
• B. \(|z_1||z_2|\sin\theta\)
• C. \(|z_1||z_2|\cos\theta\)
• D. \(|z_1||z_2|\tan\theta\)

Hint:

The dot product of two complex numbers \(z_1\) and \(z_2\) can also be computed as:
\[ z_1 \cdot z_2 = \text{Re}(z_1 \bar{z_2}) \] Since \(z_1 \bar{z_2} = (|z_1|e^{i\theta_1})(|z_2|e^{-i\theta_2}) = |z_1||z_2| e^{i(\theta_1 - \theta_2)}\), taking the real part gives:
\[ \text{Re}(z_1 \bar{z_2}) = |z_1||z_2|\cos(\theta_1 - \theta_2) = |z_1||z_2|\cos\theta \]

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
This question asks for the dot product of two complex numbers \(z_1\) and \(z_2\) when represented as vectors in the 2D Argand plane.
Key Formula or Approach:
For two vectors \(\vec{u}\) and \(\mathbf{v}\) in \(\mathbb{R}^2\), the dot product is defined as:
\[ \vec{u} \cdot \vec{v} = \|\vec{u}\| \|\vec{v}\| \cos\theta \] where \(\theta\) is the angle between the two vectors.
For complex numbers \(z_1\) and \(z_2\), their vector magnitudes are equal to their moduli \(|z_1|\) and \(|z_2|\).

Step 2: Detailed Explanation:

• Let \(z_1 = x_1 + i y_1\) and \(z_2 = x_2 + i y_2\).
In vector form, they are represented as \(\vec{z_1} = (x_1, y_1)\) and \(\vec{z_2} = (x_2, y_2)\).

• The standard dot product of these vectors is:
\[ \vec{z_1} \cdot \vec{z_2} = x_1 x_2 + y_1 y_2 \]

• We can also express this in terms of polar coordinates:
Let \(z_1 = r_1 e^{i\theta_1}\) and \(z_2 = r_2 e^{i\theta_2}\).
Then \(x_1 = r_1 \cos\theta_1, y_1 = r_1 \sin\theta_1\) and \(x_2 = r_2 \cos\theta_2, y_2 = r_2 \sin\theta_2\).

• Substituting these into the dot product formula:
\[ \vec{z_1} \cdot \vec{z_2} = (r_1 \cos\theta_1)(r_2 \cos\theta_2) + (r_1 \sin\theta_1)(r_2 \sin\theta_2) \] \[ \vec{z_1} \cdot \vec{z_2} = r_1 r_2 (\cos\theta_1 \cos\theta_2 + \sin\theta_1 \sin\theta_2) \] Using the trigonometric identity \(\cos(\theta_1 - \theta_2) = \cos\theta_1 \cos\theta_2 + \sin\theta_1 \sin\theta_2\):
\[ \vec{z_1} \cdot \vec{z_2} = r_1 r_2 \cos(\theta_1 - \theta_2) \]

• Let \(\theta = \theta_1 - \theta_2\) be the angle between the two complex numbers.
Since \(r_1 = |z_1|\) and \(r_2 = |z_2|\), we get:
\[ \text{Dot Product} = |z_1| |z_2| \cos\theta \]

Step 3: Final Answer:
The dot product of \(z_1\) and \(z_2\) is \(|z_1||z_2|\cos\theta\).