Question:
Consider two complex numbers \( z_1 = 1 - i \) and \( z_2 = i \). The argument of \( z_1z_2 \) is:
Consider two complex numbers \( z_1 = 1 - i \) and \( z_2 = i \). The argument of \( z_1z_2 \) is:
Show Hint
When multiplying complex numbers, convert them to polar form for easier multiplication and determination of arguments.
Updated On: Apr 11, 2025
- 0
- \(\frac{\pi}{4}\)
- \(\frac{\pi}{2}\)
- \(\pi\)
Show Solution
Verified By Collegedunia
The Correct Option is B
Solution and Explanation
Multiply the complex numbers.
\[
z_1z_2 = (1 - i) \cdot i = i - i^2 = i + 1
\]
Step 2: Determine the argument.
The complex number \( i + 1 \) is in the first quadrant of the complex plane. The angle corresponding to a vector with equal real and imaginary parts is \(\frac{\pi}{4}\).
The complex number \( i + 1 \) is in the first quadrant of the complex plane. The angle corresponding to a vector with equal real and imaginary parts is \(\frac{\pi}{4}\).
Was this answer helpful?
0
0
Top GATE CH Engineering Mathematics Questions
- Consider a Leniar Homogeneous system AX = 0, where A is N × N matrix and X is N × 1 vector. When we solve this 0 is null matrix of N × 1. Let R be the rank matrix for a for a Non-trivial solution.
- F(x) = ex - 5x and the value of X0 = 1 then what will be the value of X1?
- The first non-zero term in the taylor series expansion of (1 - x) - e- x, about x = 0
- \(\frac{P}{Q}\) is not equal 1, \(\left(\frac{P}{Q}\right)^{\frac{P}{Q}} = P^{\frac{P}{Q} - 1}\)
- What will be the argument of \(\frac{Z_1}{Z_2}\)=?
\(Z_1 = -1 + i\)
\(Z_2 = 2i\)
View More Questions
Top GATE CH Questions
- Consider a Leniar Homogeneous system AX = 0, where A is N × N matrix and X is N × 1 vector. When we solve this 0 is null matrix of N × 1. Let R be the rank matrix for a for a Non-trivial solution.
- Which are true for Petroleum / Petrochemicals?
- F(x) = ex - 5x and the value of X0 = 1 then what will be the value of X1?
- The first non-zero term in the taylor series expansion of (1 - x) - e- x, about x = 0
- \(\frac{P}{Q}\) is not equal 1, \(\left(\frac{P}{Q}\right)^{\frac{P}{Q}} = P^{\frac{P}{Q} - 1}\)
View More Questions