Question

Let \(Z_1 = r_1(\cos \theta_1 + i\sin \theta_1)\) and \(Z_2 = r_2(\cos \theta_2 + i\sin \theta_2)\) be two complex numbers, then value of \(Z_1Z_2\) is

• A. \(r_1r_2[\cos(\theta_1\theta_2) + i\sin(\theta_1\theta_2)]\)
• B. \(r_1r_2[\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)]\)
• C. \(r_1r_2[\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2)]\)
• D. \(\frac{r_1}{r_2}[\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)]\)

Hint:

When multiplying complex numbers in polar form:
- Multiply their magnitudes: \( r_{\text{new}} = r_1 \cdot r_2 \).
- Add their arguments (angles): \( \theta_{\text{new}} = \theta_1 + \theta_2 \).
This simple rule saves you from performing long algebraic expansions.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
This question tests the multiplication of complex numbers represented in polar form.
It can be solved using standard trigonometric identity formulas or complex exponential form (Euler's form).

Step 2: Detailed Explanation:
Let us multiply the two complex numbers algebraically:
\[ Z_1 = r_1(\cos \theta_1 + i\sin \theta_1) \] \[ Z_2 = r_2(\cos \theta_2 + i\sin \theta_2) \] The product \( Z_1 Z_2 \) is:
\[ Z_1 Z_2 = [r_1(\cos \theta_1 + i\sin \theta_1)] \cdot [r_2(\cos \theta_2 + i\sin \theta_2)] \] \[ Z_1 Z_2 = r_1 r_2 (\cos \theta_1 + i\sin \theta_1)(\cos \theta_2 + i\sin \theta_2) \] Let us expand the product inside the parentheses:
\[ (\cos \theta_1 + i\sin \theta_1)(\cos \theta_2 + i\sin \theta_2) = \cos \theta_1 \cos \theta_2 + i \cos \theta_1 \sin \theta_2 + i \sin \theta_1 \cos \theta_2 + i^2 \sin \theta_1 \sin \theta_2 \] Since \( i^2 = -1 \), substitute this value and group the real and imaginary parts:
\[ = (\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2) + i(\sin \theta_1 \cos \theta_2 + \cos \theta_1 \sin \theta_2) \] Using the standard trigonometric sum formulas:
\[ \cos(\theta_1 + \theta_2) = \cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2 \] \[ \sin(\theta_1 + \theta_2) = \sin \theta_1 \cos \theta_2 + \cos \theta_1 \sin \theta_2 \] Substitute these identities back into the expression:
\[ = \cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2) \] Therefore, the final product is:
\[ Z_1 Z_2 = r_1 r_2 [\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)] \] Let us verify using Euler's exponential form, which is much faster:
\[ Z_1 = r_1 e^{i\theta_1} \quad \text{and} \quad Z_2 = r_2 e^{i\theta_2} \] \[ Z_1 Z_2 = (r_1 e^{i\theta_1})(r_2 e^{i\theta_2}) = r_1 r_2 e^{i(\theta_1 + \theta_2)} \] Convert the exponential term back to polar form:
\[ Z_1 Z_2 = r_1 r_2 [\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)] \] This matches Option (B).

Step 3: Final Answer:
The product is \( r_1 r_2 [\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)] \).
Therefore, the correct choice is Option (B).