Question

The point \(z=\frac{1}{\sqrt{2}}(1+i)\) in the complex plane is rotated about the origin through an angle \(\frac{\pi}{4}\) in the clockwise direction, then the new position of \(z\) is

• A. \(2\)
• B. \(1\)
• C. \(\frac{1}{\sqrt{2}}(1-i)\)
• D. \(\frac{1}{2}(1-i)\)
• E. \(1-i\)

Hint:

In the complex plane, rotation about the origin changes only the argument, not the modulus. Clockwise rotation means subtract the angle from the argument.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Rotating a complex number \(z\) about the origin by an angle \(\theta\) in the counter-clockwise direction is equivalent to multiplying \(z\) by the complex number \(e^{i\theta} = \cos\theta + i\sin\theta\). A rotation in the clockwise direction corresponds to a negative angle.
Step 2: Key Formula or Approach:
Let the new position of \(z\) be \(z'\). The rotation is by \(\theta = -\frac{\pi}{4}\) (clockwise).
The formula for the new position is:
\[ z' = z \cdot e^{i\theta} = z \cdot (\cos\theta + i\sin\theta) \] Step 3: Detailed Explanation:
Method 1: Using Rectangular Coordinates
The rotation factor for a clockwise rotation by \(\frac{\pi}{4}\) is:
\[ e^{-i\pi/4} = \cos(-\frac{\pi}{4}) + i\sin(-\frac{\pi}{4}) = \cos(\frac{\pi}{4}) - i\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} - i\frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}}(1-i) \] Now, multiply the original complex number \(z\) by this factor:
\[ z' = \left(\frac{1}{\sqrt{2}}(1+i)\right) \cdot \left(\frac{1}{\sqrt{2}}(1-i)\right) \] \[ z' = \frac{1}{(\sqrt{2})^2} (1+i)(1-i) \] Using the difference of squares formula, \((a+b)(a-b) = a^2 - b^2\):
\[ z' = \frac{1}{2} (1^2 - i^2) \] Since \(i^2 = -1\):
\[ z' = \frac{1}{2} (1 - (-1)) = \frac{1}{2}(2) = 1 \] Method 2: Using Polar Form
First, convert \(z\) to its polar form \(re^{i\alpha}\).
The modulus is \(|z| = |\frac{1}{\sqrt{2}}(1+i)| = \frac{1}{\sqrt{2}}|1+i| = \frac{1}{\sqrt{2}}\sqrt{1^2+1^2} = \frac{1}{\sqrt{2}}\sqrt{2} = 1\).
The argument is \(\alpha = \arg(1+i) = \arctan(\frac{1}{1}) = \frac{\pi}{4}\).
So, \(z = 1 \cdot e^{i\pi/4}\).
To rotate by \(-\frac{\pi}{4}\), we multiply by \(e^{-i\pi/4}\):
\[ z' = z \cdot e^{-i\pi/4} = (e^{i\pi/4}) \cdot (e^{-i\pi/4}) = e^{i(\pi/4 - \pi/4)} = e^{i0} \] \[ z' = \cos(0) + i\sin(0) = 1 + 0i = 1 \] Step 4: Final Answer:
The new position of z after the rotation is 1. Therefore, option (B) is the correct answer.