Question

If $a = cos\theta + i \cdot sin\theta$ then $1+a^{2}$ is:

• A. $2 \cdot cos\theta(cos\theta + i \cdot sin\theta)$
• B. $2 \cdot cos\theta(cos\theta - i \cdot sin\theta)$
• C. $2 \cdot sin\theta(cos\theta + i \cdot sin\theta)$
• D. $cos\theta(cos\theta + i \cdot sin\theta)$

Hint:

Always look to use the identity $1 + cos(2\theta) = 2 \cdot cos^2\theta$ to simplify expressions involving $1 + e^{i2\theta}$.

Answer 1

The Correct Option is A

Concept: According to De Moivre's Theorem, for any complex number in polar form $a = e^{i\theta} = cos\theta + i \cdot sin\theta$, the square is $a^2 = e^{i2\theta} = cos(2\theta) + i \cdot sin(2\theta)$. We then simplify the expression $1+a^2$ using trigonometric identities.

Step 1: Substitute $a^2$ and simplify.
Given $a = cos\theta + i \cdot sin\theta$: \[ 1 + a^2 = 1 + (cos\theta + i \cdot sin\theta)^2 \] \[ 1 + a^2 = 1 + cos(2\theta) + i \cdot sin(2\theta) \]

Step 2: Apply double-angle identities.
Recall that $1 + cos(2\theta) = 2 \cdot cos^2\theta$ and $sin(2\theta) = 2 \cdot sin\theta \cdot cos\theta$: \[ 1 + a^2 = 2 \cdot cos^2\theta + i(2 \cdot sin\theta \cdot cos\theta) \]

Step 3: Factor out the common term.
Factor out $2 \cdot cos\theta$: \[ 1 + a^2 = 2 \cdot cos\theta(cos\theta + i \cdot sin\theta) \] This matches option (1).