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: Write the given complex number in trigonometric form.
We are given:
\[ z=\frac{1}{\sqrt{2}}(1+i)=\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}} \] Its modulus is:
\[ |z|=\sqrt{\left(\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{\sqrt{2}}\right)^2} =\sqrt{\frac{1}{2}+\frac{1}{2}}=1 \] So \(z\) lies on the unit circle.

Step 2: Find its argument.
Since the real part and imaginary part are both positive and equal, the point lies in the first quadrant at angle:
\[ \arg(z)=\frac{\pi}{4} \] Therefore,
\[ z=\cos\frac{\pi}{4}+i\sin\frac{\pi}{4} \]

Step 3: Understand clockwise rotation.
A clockwise rotation by angle \(\frac{\pi}{4}\) means the argument decreases by \(\frac{\pi}{4}\). So the new argument becomes:
\[ \frac{\pi}{4}-\frac{\pi}{4}=0 \]

Step 4: Use the rotation rule in complex numbers.
Rotation through angle \(\theta\) clockwise about the origin corresponds to multiplication by:
\[ \cos(-\theta)+i\sin(-\theta)=e^{-i\theta} \] So here the new point is:
\[ z' = z\left(\cos\frac{\pi}{4}-i\sin\frac{\pi}{4}\right) \]

Step 5: Apply the angle change directly.
Since the modulus remains unchanged and only the argument changes, we get:
\[ z'=\cos 0+i\sin 0 \] \[ z'=1 \]

Step 6: Verify algebraically.
Also, \[ z=\frac{1}{\sqrt{2}}(1+i) \] and multiplying by \[ \frac{1}{\sqrt{2}}(1-i) \] gives: \[ z'=\frac{1}{2}(1+i)(1-i) \] \[ =\frac{1}{2}(1-i^2)=\frac{1}{2}(1+1)=1 \]

Step 7: Match with the options.
Hence the new position of \(z\) is:
\[ 1 \] This matches option \((2)\).