Question

If \[ z=\cos\theta+i\sin\theta \] then \[ z^r+(\overline{z})^r= \] is equal to:

• A. \(\cos r\theta\)
• B. \(2\cos r\theta\)
• C. \(\sin r\theta\)
• D. \(2\sin r\theta\)

Hint:

If \(z=\cos\theta+i\sin\theta\), then by De Moivre's theorem: \[ z^r=\cos r\theta+i\sin r\theta \] and \[ (\overline{z})^r=\cos r\theta-i\sin r\theta \] Their sum is always \(2\cos r\theta\).

Answer 1

The Correct Option is B

Step 1: Write \(z\) in trigonometric form.
Given \[ z=\cos\theta+i\sin\theta \] By De Moivre's theorem, \[ z^r=(\cos\theta+i\sin\theta)^r \] Therefore, \[ z^r=\cos r\theta+i\sin r\theta \]

Step 2: Find the conjugate of \(z\).
Since \[ z=\cos\theta+i\sin\theta, \] we have \[ \overline{z}=\cos\theta-i\sin\theta \] Now, \[ (\overline{z})^r=(\cos\theta-i\sin\theta)^r \] Using De Moivre's theorem, \[ (\overline{z})^r=\cos r\theta-i\sin r\theta \]

Step 3: Add \(z^r\) and \((\overline{z})^r\).
Now, \[ z^r+(\overline{z})^r = (\cos r\theta+i\sin r\theta)+(\cos r\theta-i\sin r\theta) \] The imaginary terms cancel out: \[ i\sin r\theta-i\sin r\theta=0 \] So, \[ z^r+(\overline{z})^r=2\cos r\theta \]

Step 4: Final conclusion.
Therefore, \[ \boxed{2\cos r\theta} \]